fantasticblazedinosaur-blog

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 7 Quiz with Formulas – Latest

DeVry MATH 221 Week 7 Quiz with Formulas – Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3377&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

A mechanic sells     a brand of automobile tire that has a life expectancy that is normally     distributed, with a mean life of 35,000 miles and a standard deviation of     2800 miles. He wants to give a guarantee for free replacement of tires     that don’t wear well. How should he word his guarantee if he is willing to     replace approximately 10% of the tires?

Tires that wear out by 31,412 miles will be replaced free of charge.

Hint:

Here we need a value of X say K such that P (X < K) = 0.10

We know P (Z < -1.282) = 0.10

Thus from Z score of K we have,

Solving for K we have K = 31412

  Find the indicated     z-score shown in the graph to the right.

The z-score is -1.18

Hint:

From standard normal table, P (Z < -1.18) = 0.1190

Therefore, z = -1.18

  A researcher     wishes to estimate, with 95% confidence, the amount of adults who have     high-speed internet access. Her estimate must be accurate within 4% of the     true proportion.     a) find the minimum sample size needed, using a prior study that found     that 32% of the respondents said they have high-speed internet access.

b) no     preliminary estimate is available. Find the minimum sample size needed.

a) = 523 b) = 601

Hint:

 (a)

The sample size is given by where p = 0.32

Given that E = 4% = 0.04, = 1.959963985

Therefore, sample size,

That is, n >522.4384

Hence the minimum sample size required is n = 523

(b)

The sample size is given by where p = 0.50

Given that E = 4% = 0.04, = 1.959963985

Therefore, sample size,

That is, n >600.2279

Hence the minimum sample size required is n = 601

  The total     cholesterol levels of a sample of men aged 35-44 are normally distributed     with a mean of 221 milligrams per deciliter and a standard deviation of     37.7 milligrams per deciliter.     (a) what percent of men have a total cholesterol level less than 228     milligrams per deciliter of blood?     (b) if 251 men in the 35-44 age group are randomly selected, about how     many would you expect to have a total cholesterol level greater than 264     milligrams per deciliter of blood?

a) = 57.37% b) = 32

Hint:

(a) P (X < 228) = = P (Z <0.1857) = 0.5737 = 57.37%

(b) P (X > 264) = = P (Z >1.1406) = 0.127

Hence the number = 251 * 0.127 = 32

Find the z-score     that has a 12.1% of the distribution’s area to it’s left.     Answer = -1.17

Hint:

We need the k such that P (Z < k) = 0.121

From standard normal table, P (Z < -1.17) = 0.121

Therefore k = -1.17

A doctor wants     to estimate the HDL cholesterol of all 20-29 year old females. How many     subjects are needed to estimate the HDL cholesterol within 2 points with     99% confidence assuming σ = 18.1? suppose the doctor would be content with     90% confidence. How does the decrease in confidence affect the sample size     required?

99% = 554 subjects 90% = 222 subjects

how does the decrease in confidence affect the sample size required? Answer: The lower the confidence% level the smaller the sample size.

Hint:

The minimum sample size is given by the formula

Please see the excel spreadsheet for calculations.

  Use a table of     cumulative areas under the normal curve to find the z-score that     corresponds to the given cumulative area. If the area is not in the table,     use the entry closest to the area. If the area is halfway between two     entries, use the z-score halfway between the corresponding z-scores. If     convenient, use technology to find the z-score. 0.054

Answer: -1.607

Hint:

We need the k such that P (Z < k) = 0.054

From standard normal table, P (Z < -1.607) = 0.121

Therefore k = -1.607

   In a survey of     3076 adults, 1492 say they have started paying bills online in the last     year.     Construct a 99% confidence interval for the population proportion.     Interpret the results.

Answers: 462, 0.508

With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between the endpoints of the given confidence interval.

Hint:

99% Confidence Interval for proportion is given by

Please see the excel spreadsheet for calculation.

  Assume the     random variable x is normally distributed with mean u = 89 and standard     deviation o = 4. Find the indicated probability. P(76<x<82)     Answer: 0395

Hint:

P (76 <X <82) = = P (-3.25 <Z < -1.75) = 0.0395

  Find the margin     of error for the given values of c,s, and n.     c = .90, s = 3.1, n =49     Answer: 728

Hint:

Margin of error =  = 1.645 * (3.1/√49) = 0.728

Find the     critical value Tc for the confidence level c = .90 and sample size n = 29.     Tc = 701

Hint:

Degrees of freedom = n – 1 = 29 – 1 = 28

From Student’s t distribution table, critical value Tc with d.f. 28 at the confidence level 0.90 is given by,

Tc = 1.701

The mean height     of women in a country (ages 20-29) is 63.9 inches. A random sample of 65     women in this age group is selected. What is the probability that the mean     height for the sample is greater than 65 inches? Assume o = 2.91     Answer = 0012

Hint:

P (>65) == P (Z >3.0476) = 0.0012

  A survey was     conducted to measure the height of men. In the survey, respondents were     grouped by age. In the 20-29 age group, the heights were normally     distributed with a mean of 67.9 inches and a standard deviation of 3.0     inches. A study participant is randomly selected. Complete parts (A)     through (C).     (a) the probability that his height is less than 68 inches.     Answer: 5133     (b) the probability that his height is between 68-71 inches.     Answer: 0.3360     (c) the probability that his height is more than 71 inches.     Answer: 0.1507

Hint:

(a) P (X <68) = = P (Z < 0.03333) = 0.5133

(b) P (68 <X <71) = = P (0.0333 <Z < 1.0333) = 0.3360

(c) P (X >71) = = P (Z > 1.03333) = 0.1507

  For the standard     normal distribution shown on the right, find the probability of z     occurring in the region.

Probability = .6950

Hint:

P (Z < 0.51) = 0.6950

Find the     indicated probability using the standard normal distribution.     (P – 1.35 < z < 1.35) = 8230

Hint:

(P – 1.35 < z < 1.35) = P (z < 1.35) – P (z < -1.35)

= 0.9115 – 0.0885

= 0.8230

  Find the margin     of error for the given values of c, s, n.     c = 0.98, s = 5, n = 6     Answer: 9

Hint:

Margin of error = = 3.3649 * (5/√6) = 6.9

The systolic     blood pressures of a sample of adults are normally distributed with a mean     pressure of 115 millimeters of mercury and a standard deviation of 3.6     millimeters of mercury. The systolic blood pressures of four adults     selected at random are 122 millimeters of mercury, 113 millimeters of     mercury, 106 millimeters of mercury, and 128 millimeters of mercury. The     graph of the standard normal distribution is below. Complete parts a – c.

  Match the values     with the letters a/b/c/d.     A = 106     B = 113     C = 122     D = 128

Find the     z-scores that corresponds to each value     A = -2.50     B = -0.56     C = 1.94     D = 3.61

Hint:

A = (X – μ)/σ = (106 – 115)/3.6 = -2.50

B = (X – μ)/σ = (113 – 115)/3.6 = -0.56

C = (X – μ)/σ = (122 – 115)/3.6 = 1.94

D = (X – μ)/σ = (128 – 115)/3.6 = 3.61

Determine if any     of the values are unusual, and classify them as either unusual or very     unusual.

Answer: The unusual value(s) is/are 106. The very unusual value(s) is/are 128.

Hint:

106 is unusual, since the z-score is outside ±2. 128 is very unusual, since the z-score is outside ±3.

  You are given a     sample mean and standard deviation. Use this information to construct the     90% and 95% confidence intervals for the population mean. Interpret the     results and compare the widths of the confidence intervals. If convenient,     use technology to construct the confidence intervals. A random sample of     60 home theater systems has a mean price of $134.00 and a standard     deviation is $19.60

The 90% confidence interval is (84), (138.16). The 95% confidence interval is (129.04, (138.96).

Interpret the results.

Answer: With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than the 90%.

The monthly     incomes for 12 randomly select people, each with a bachelor’s degree in     economics, are shown on the right. Assume the population is normally     distributed.

Mean = 4263.3 Standard Deviation = 260.1 99% confidence interval = 4030.1, 4496.5

  What is the     total area under the normal curve?     Answer = 1

A population has     a mean u = 82 and a standard deviation o = 36. Find the mean and standard     deviation of a sampling distribution of sample means with sample size n =     81     82

4

The amounts of     time employees at a large corporation work each day are normally     distributed, with a mean of 7.4 hours and a standard deviation of 0.38     hour. Random sample of size 25 and 37 are drawn from the population and     the mean of each sample is determined. What happens to the mean and the     standard deviation of the distribution of sample means as the size of the     sample increases?

Mean of distribution = 4 Standard deviation of distribution = 0.08

If the sample size is n = 37, find the mean and standard deviation.

Mean = 7.4 Standard deviation = 0.06

What happens to the mean and standard deviation of the distribution of sample means as the size of the sample increases? Answer: The mean stays the same, the but standard deviation decreases.

Assume a member     is selected at random from the population represented by the graph. Find     the probability that the member selected at random is from the shaded area     of the graph.

Answer = 0.3038

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 7 Homework - Latest

DeVry MATH 221 Week 7 Homework - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3332&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

DeVry MATH 221     Week 7 Homework - Latest 2015

    Use the given     statement to represent a claim. Write it’s complement and state which is     Ho and which is Ha.

u> 635

Find the complement of the claim. u < 635

   A null and     alternative hypothesis are given. Determine whether the hypothesis test is     left-tailed, right tailed, or two-tailed.

What type of test is being conducted in this problem?

Answer: Right-tailed test

  Write the null     and alternative hypotheses. Identify which is the claim.     A light bulb manufacturer claims that the mean life of a certain type of     light bulb is more than 700 hours.

Identify which is the claim.

Answer: The alternative hypothesis Ha is the claim.

  Write the null     and alternative hypotheses. Identify which is the claim. The standard     deviation of the base price of a certain type of car is at least $1010.

Identify which is the claim.

Answer: The null hypothesis Ho is the claim.

  More than 11% of     all homeowners have a home security alarm. Determine whether the     hypothesis for this claim is left-tailed, right-tailed, or two-tailed.     Explain your reasoning.     Answer: The hypothesis test is right-tailed because the alternative     hypothesis contains >

A film developer     claims that the mean number of pictures developed for a camera with 22     exposures is less than 17. If a hypothesis test is performed, how should     interpret a decision that (a) rejects the null hypothesis and (B) fails to     reject the null hypothesis?

A = There is enough evidence to support the claim that the mean number of pictures developed for a camera with 22 exposures is less than 17. B = There is not enough evidence to support the claim that the mean number of pictures developed for a camera with 22 exposures is less than 17.

  Find the P-value     for the indicated hypothesis test with the given test statistic, z. Decide     whether to reject Ho for the given level of significance a.     Two-tailed test with test statistic z = -2.08 and a = 0.04     P-Value = 0376     Conclusion = Reject Ho

Find the     critical z values. Assume that the normal distribution applies.     Right-tailed test, a = 08.     Z = 1.41

Find the     critical value(s) for a left-tailed z-test with a = 0.01. Include a graph     with your answer.     Critical Value = -2.33

Graph:

Test the claim     about the population mean, u, at the given level of significance using the     given sample statistics.     Claim u = 50, a = 0.08, sample statistics: x = 49.2, s = 3.56, n = 80

  Standardized test statistic = 2.01 Critical Values = 1.75 Reject Ho. At the 8% significance level, there Is enough evidence to reject the claim.

Test the claim     about the population mean, u, at the given level of significance using the     given sample statistics.     Claim u = 5000, a = 0.05. Sample statistics x = 4800, s = 323, n = 46.

Standardized test statistic = -4.20 Critical Values = 1.96 Reject Ho. At the 5% significance level, there is enough evidence to support the claim.

A random sample     of 85 eight grade students’ score on a national mathematics assessment     test has a mean score of 275 with a standard deviation of 33. This test     result prompts a state school administrator to declare that the mean score     for the state’s eighth graders on this exam is more than 270. At a = 0.03,     is there enough evidence to support the administrators claim? Compare     parts A – E.

Z = 1.40 Area = 0.919 P Value = 0.081 Reject Ho At the 3% significance level, there is not enough evidence to support the administrator’s claim that the mean score for the state’s 8th graders on the exam is more than 270.

  A company that     makes cola drinks states that the mean caffeine content per 12-ounce     bottle of cola is 45 milligrams. You want to test this claim. During your     tests, you find that a random sample of thirty 12-ounce bottles of the     cola has a mean caffeine content of 45.5 milligrams with a standard     deviation of 6.1 milligrams. At a = 0.08, can you reject the company’s     claim?

The critical values are = 1.75

z = 0.45 Since z is not in the rejection region, fail to reject the null hypothesis. At the 8% significance level, there is not enough evidence to reject the company’s claim that the mean caffeine content per 12-ounce bottle of cola is equal to 45 milligrams.

  A light bulb     manufacturer guarantees that the mean life of a certain type of light bulb     is at least 975 hours. A random sample of 72 light bulbs has a mean life     of 954 hours with a standard deviation of 85 hours. Do you have enough     evidence to reject the manufacturer’s claim? Use a = 0.04.

Zo = -1.75 Z = -2.10 Reject Ho. There is sufficient evidence to reject the claim that mean bulb life is at least 975 hours.

  An     environmentalist estimates that the mean waste recycled by adults in the     country is more than 1 pound per person per day. You want to this test     claim. You find that the mean waste recycled per person per day for a     random sample of 12 adults in the country is 1.4 pounds and the standard     deviation is 0.3 pound. At a = 0.10, can you support the claim? Assume the     population is normally distributed.

To = 1.363

T = 4.62 Reject Ho because the standardized test statistic is in the rejection region.

A county is     considering raising the speed limit on a road because they claim that the     mean speed of vehicles is greater than 40 miles per hour. A random sample     of 25 vehicles has a mean speed of 45 miles per hour and a standard     deviation of 5.4 miles per hour. At a = 0.10, do you have enough evidence     to support the county’s claim? Complete parts A – D.

T = 4.63 P- Value = 0.000 Reject Ho because the P-value is less than the significance level, 0.10. There is sufficient evidence to support the county’s claim that the mean speed of vehicles is greater than 40 miles per hour.

A traveled     association claims that the mean daily meal cost for two adults traveling     together on vacation is $100. A random sample of 20 such groups of adults     has a mean daily meal cost of $95 and a standard deviation of $4.50. Is     there enough evidence to reject the claim at a = 0.1? complete parts A – D.

T = -4.97 P Value = 0.000 Reject Ho because the P-value is less than the significance level, 0.1. There is sufficient evidence at the 10% level of significance to reject the travel association’s claim that the mean daily meal cost for two adults traveling together on vacation is $100.

Decide whether     the normal sampling distribution can be used. If it can be used, test the     claim about the population proportion p at the given level of significance     a using given sample statistics.     Claim p = 0.23, a = 0.10, sample statistics: p = 0.18, n = 150.

Can the normal sampling distribution be used? Answer: Yes, because both np and nq are greater than or equal to 5.

The critical values are = -1.64, 1.64 z = -1.46 What is the result of the test? Answer: Fail to reject Ho. The data do not provide sufficient evidence to support the claim.

  (a) write the     claim mathematically and identify Ho and Ha. (b) find the critical     value(s) and identify the rejection region(s). find the standardized test     statistic. (d) decide whether to reject or fail to reject the null     hypothesis. An environmental agency recently claimed more than 25% of     consumers have stopped buying a certain product because of environmental     concerns. In a random sample of 1000 customers, you find 40% have stopped     buying the product. At a=0.03, do you have enough evidence to support the     claim?

Zo = 1.88

Z = 10.95 Reject Ho. There is enough evidence to support the claim.

A humane society     claims that less than 36% of U.S. households own a dog. In a random sample     of 406 U.S. households, 154 say they own a dog. At a = 0.10, is there     enough evidence to support the society’s claim?     (a) write the claim mathematically and identify Ho and Ha. (b) find the     critical value(s) and identify the rejection region(s). (c) find the     standardized test statistic. (d) decide whether to reject or fail to     reject the null hypothesis, and € interpret the decision in the context of     the original claim.

Critical Values = -1.28

A = 0.60 Fail to reject Ho There is not enough evidence to support the claim that less than 36% of U.S. households own a dog.

   ":"qui���w#

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 7 DQ Rejection Region - Latest

DeVry MATH 221 Week 7 DQ Rejection Region - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3373&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Week 7 DQ Rejection Region

How is the rejection region defined and how is that related to the z-score and the p value? When do you reject or fail to reject the null hypothesis? Why do you think statisticians are asked to complete hypothesis testing? Can you think of examples in courts, in medicine, or in your area?

   \"sh���/`-

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 6 Homework - Latest

DeVry MATH 221 Week 6 Homework - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3328&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Homework Week 6 -2015

 1  Given the same sample statistics, which level of confidence would produce the widest confidence interval?

Choose the correct answer below.

  90%

98%

99%

95%

  Given the same     sample statistics, which level of confidence would produce the widest     confidence interval?

Choose the correct answer below.

99%

95%

98%

90%

 Use the values on the number line to find the sampling error.

The sampling error is 1.56

  Use the values     on the number line to find the sampling error.

  The sampling error is 1.69

 3  Find the margin of error for the given values of c, s, and n.

C=0.90, s=3.4, n=81

 E= 0.621 (Round to three decimal places as needed)

  Find the margin     of error for the given values of c, s, and n.

C=0.90, s=3.6, n=81

 E= .658 (Round to three decimal places as needed)

   Construct the     confidence interval for the population mean µ.

A 98% confidence interval for µ is (5.95,6.45) (Round to two decimal places as needed)

 4  Construct the confidence interval for the population meanµ.

 A 98% confidence interval for µ is (7.04,7.16),  (Round to two decimal places as needed)

  Construct the     confidence interval for the population mean µ.

 A 90% confidence interval for µ is (15.5,17.9). (Round to one decimal place as needed)

  5  Construct the confidence interval for the population mean µ.

 A 98% confidence interval for µ is (15.1,16.5). (Round to one decimal place as needed)

  6  Use the confidence interval to find the estimated margin of error.  Then find the sample mean.

A biologist reports a confidence interval of (4.3,5.1) when estimating the mean height (in centimeters) of a sample of seedlings.

 The estimated margin of error is 0.4.

The sample mean is 4.7.

   Use the     confidence interval to find the estimated margin of error. Then find the     sample mean.

A biologist reports a confidence interval of (1.6,3.2) when estimating the mean height (in centimeters) of a sample of seedlings.

 The estimated margin of error is .8.

The sample mean is 2.4.

 7  Find the minimum sample size n needed to estimate µ for the given values of c, s, and E

 C=0.90, s=7.7, and E=1

 Assume that a preliminary sample has at least 30 members

N=161 (Round up to the nearest whole number)

  Find the minimum     sample size n needed to estimate µ for the given values of c, s, and E.

C=0.95, s= 9.3, and E=1

Assume that a preliminary sample has at least 30 members

N=333 (Round up to the nearest whole number)

 8You are given the sample mean and the sample standard deviation.  Use this information to construct the 90% and 95% confidence intervals for the population mean.  Interpret the results and compare the widths of the confidence intervals.  If convenient, use technology to construct the confidence intervals.

A random sample of 55 home theater system has a mean price of $135.00 and a standard deviation is $15.70.

 Construct a 90% confidence interval for the population mean.

The 90% confidence interval is (131.52,138.48) (Round to two decimal places as needed)

 Construct a 95% confidence interval for the population mean.

The 95% confidence interval is (130.85,139.15) (Round to two decimal places as needed)

 Interpret the results. Choose the correct answer below.

With 90%     confidence, it can be said that the population mean price lies in the     first interval. With 95% confidence, it can be said that the population     mean price lies in the second interval. The 95% confidence interval is     wider than 90%.     B. With 90% confidence, it can be said that the sample mean price lies in     the first interval. With 95% confidence, it can be said that the sample     mean price lies in the second interval. The 95% confidence interval is wider     than the 90%.     C. With 90% confidence, it can be said that the population mean price lies     in the first interval. With 95%, confidence, it can be said that the     population mean price lies in the second interval. The 95% confidence     interval is narrower than 90%.

     You are given     the sample mean and the sample standard deviation. Use this information to     construct the 90% and 95% confidence intervals for the population     mean.  Interpret the results and compare the widths of the confidence     intervals.  If convenient, use technology to construct the confidence     intervals.

A random sample of 60 home theater system has a mean price of $115.00 and a standard deviation is $15.10.

 Construct a 90% confidence interval for the population mean.

The 90% confidence interval is (111.79,118.207) (Round to two decimal places as needed)

 Construct a 95% confidence interval for the population mean.

The 95% confidence interval is (111.18,118.82) (Round to two decimal places as needed)

 Interpret the results. Choose the correct answer below. A. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95% confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is wider than 90%. B. With 90% confidence, it can be said that the sample mean price lies in the first interval. With 95% confidence, it can be said that the sample mean price lies in the second interval. The 95% confidence interval is wider than the 90%. C. With 90% confidence, it can be said that the population mean price lies in the first interval. With 95%, confidence, it can be said that the population mean price lies in the second interval. The 95% confidence interval is narrower than 90%.

  You are given     the sample mean and the sample standard deviation. Use this information to     construct the 90% and 95% confidence intervals for the population mean.     Which interval is wider? If convenient, use technology to construct the     confidence intervals.     A random sample of 31 gas grills has a mean price of $642.90 and a     standard deviation of $58.40.

The 90% confidence interval is (6,660.2) (round to one decimal place as needed)

The 95% confidence interval is (622.3,663.5) (round to one decimal place as needed)

 Which interval is wider?  Choose the correct answer below.

  The 90%     confidence interval

The 95%     confidence interval

  You are given     the sample mean and the sample standard deviation. Use this information to     construct the 90% and 95% confidence intervals for the population mean.     Which interval is wider? If convenient, use technology to construct the     confidence intervals.

A random sample of 33 eight-ounce servings of different juice drinks has a mean of 93.5 calories and a standard deviation of 41.5 calories.

The 90% confidence interval is (81.6,105.4).(Round to 1 decimal place as needed.)

The 95% confidence interval is (79.3,107.7).(Round to 1 decimal place as needed.)

 Which interval is wider?

The 95%     confidence interval

The 90%     confidence interval

  People were     polled on how many books they read the previous year. How many subjects     are needed to estimate the number of books read the previous year within     one book with 90% confidence? Initial survey results indicate that σ=11.7     books

 A 90% confidence level requires 371 subjects. (Round up to the nearest whole # as needed)

  A doctor wants     to estimate the HDL cholesterol of all 20-to 29-year-old females. How many     subjects are needed to estimate the HDL cholesterol within 2 points with     99% confidence assuming σ=15.4? Suppose that the doctor would be content     with 95% confidence. How does the decrease in confidence affect the sample     size required?

A 99% confidence level requires 394 subjects. (Round up to the nearest whole number as needed)

 A 95% confidence level requires 228 subjects. (Round up to the nearest whole number as needed)

 How does the decrease in confidence affect the sample size required?

  The sample size     is the same for all levels of confidence

The lower the     confidence level the larger the sample size

The lower the     confidence level the smaller the sample size

  Construct the     indicated confidence interval for the population mean µ using (a) a     t-distribution. (b) if you had incorrectly used a normal distribution,     which interval would be wider?

  The 95%     confidence interval using a t-distribution is (3,16.9)     (round to one decimal place as needed.)

If you had     incorrectly used a normal distribution, which interval would be wider?

  The     t-distribution has the wider interval

The normal     distribution has the wider interval

  In the following     situation, assume the random variable is normally distributed and use a     normal distribution or a t-distribution to construct a 90% confidence     interval for population mean. If convenient, use technology to construct     the confidence interval.

(a) In a random sample of 10 adults from a nearby county, the mean waste generated per person per day was 4.65 pounds and the standard deviation was 1.48 pounds.

Repeat part (a),     assuming the same statistics came from a sample size of 450. Compare the     results.

  (a) For the sample size of 10 adults, the 90% confidence interval is (3.79,5.51) (Round to 2 decimal places as needed.)

 (b) For the sample of 450 adults, the 90% confidence interval is (4.54,4.76)

(Round to 2 decimal places as needed.)

 Choose the correct observation below

  The interval     from part (a), which uses the normal distribution, is narrower than the     interval from part (b), which uses the t-distribution.

The interval     from part (a), which uses the t-distribution, is wider than the interval     from part (b), which uses the normal distribution.

The interval     from part (a), which uses the normal distribution, is wider than the     interval from part (b), which uses the t-distribution.

The interval     from part (a), which uses the t-distribution, is narrower than the interval     from part (b), which uses the normal distribution.

  Use the given     confidence interval to find the margin of error and the sample proportion.     (0.662,0.690)     E = 014(type an integer or a decimal.)

 0.676 (Type an integer or a decimal.)

  In a survey of 633     males from 18-64, 390 say they have gone to the dentist in the past year.     Construct 90% and 95% confidence intervals for the population proportion.     Interpret the results and compare the widths of the confidence intervals.     If convenient, use technology to construct the confidence intervals.

The 90% confidence interval for the population proportion p is (.584,.648)

(Round to 3 decimal places as needed.)

 The 95% confidence interval for the population proportion p is (.578,.645)

(Round to 3 decimal places as needed.)

 Interpret your results of both confidence intervals.

  With the given     confidence, it can be said that the population proportion of males ages     18-64 who say they have gone to the dentist in the past year is between     the endpoints of the given confidence interval.

With the given     confidence, it can be said that the population proportion of males ages     18-64 who say they have gone to the dentist in the past year is not     between the endpoints of the given confidence interval.

With the given     confidence, it can be said that the sample proportion of males ages 18-64     who say they have gone to the dentist in the past year is between the     endpoints of the given confidence interval.

 Which interval is wider?

  The 90%     confidence interval

The 95%     confidence interval

  In a survey of     6000 women, 3431 say they change their nail polish once a week. Construct     a 99% confidence interval for the population proportion of women who     change their nail polish once a week.     A 99% confidence interval for the population proportion is (.556, .588)

(Round to 3 decimal places as needed)

   A researcher     wishes to estimate, with 99% confidence, the proportion of adults who have     high-speedy internet access. Her estimate must be accurate within 4% of     the true proportion.     a) Find the minimum sample size needed, using a prior study that found     that 42% of the respondents said they have a high-speedy internet access.     b) No preliminary estimate is available. Find the minimum sample size     needed.

A) What is the minimum sample size needed using a prior study that found that 42% of the respondents said they have high-speed internet access?

n = 1010 (Round up to the nearest whole # as needed.)

  B) What is the     minimum sample size needed assuming that no preliminary estimate is     available?

n =1037 (Round up to the nearest whole # as needed.)

   sty/���"^�

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 6 DQ Confidence Interval Concepts - Latest

DeVry MATH 221 Week 6 DQ Confidence Interval Concepts - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3369&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Week 6 DQ Confidence Interval Concepts

Consider the formula used for any confidence interval and the elements included in that formula. What happens to the confidence interval if you (a) increase the confidence level, (b) increase the sample size, or (c) increase the margin of error? Only consider one of these changes at a time. Explain your answer with words and by referencing the formula.

   ":{"���\`-

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 5 DQs Interpreting Normal Distributions - Latest

DeVry MATH 221 Week 5 DQs Interpreting Normal Distributions - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3361&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Week 5 DQs Interpreting Normal Distributions - 2015

 Assume that a population is normally distributed with a mean of 100 and a standard deviation of 15. Would it be unusual for the mean of a sample of 3 to be 115 or more? Why or why not?

   nex��N`-

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 4 Homework - Latest

DeVry MATH 221 Week 4 Homework - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3320&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Homework Week 4 - 2015

  The histograms     each represents part of a binomial distribution. Each distribution has the     same probability of success, p, but different numbers of trials, n.      Identify the unusual values of x in each distogram.

N = 4

N = 8

Choose the     correct answer below. Use histogram

X = 0, x = 1, x     = 2, x = 3, and x = 4

X = 3 and x = 4

X = 0 and x = 1

There are no     unusual values of x in the histogram

X = 7

X = 0, x = 1, x     = 2, x = 3, and x = 4

X = 0 and x = 1

X = 0 and x = 1

There are no     unusual values of x in the histogram

  The histograms     each represents part of a binomial distribution. Each distribution has the     same probability of success, p, but different numbers of trials, n.      Identify the unusual values of x in each distogram.

N = 4

N = 8

Choose the     correct answer below. Use histogram (a).

X = 4

X = 0, x =7, and     x = 8

X = 2

There are no     unusual values of x in the histogram

Choose the     correct answer below. Use the histogram (b)

X =0, x =7, and     x = 8

X = 4

X = 4

There are no     unusual values of x in the histogram

  About 80% of     babies born with a certain ailment recover fully. A hospital is caring for     five babies born with this ailment.  The random variable represents     the number of babies that recover fully.  Decide whether the     experiment is a binomial experiment.  If it is, identify a success,     specify the values of n, p, and q, and list the possible values of the     random variable x.

Is the experiment a binomial experiment?

Yes

No

What is a success in this experiment?

Baby doesn’t recover

Baby recovers

This is not a binomial experiment

Specify the value of n.  Select the correct choice below and fill in any answer boxes in your choice.

N = .5

This is not a binomial experiment

Specify the value of p.  Select the correct choice below and fill in any answer boxes in your choice.

P = .8

This is not a binomial experiment

Specify the value of q.  Select the correct choice below and fill in any answer boxes in your choice.

Q = .2

This is not a binomial experiment

List the possible values of the random variable x.

X = 1, 2, 3,…, 5

X = 0, 1, 2, ….4

X = 0, 1, 2, …5

This is not a binomial experiment

  About 70% of     babies born with a certain ailment recover fully. A hospital is caring for     six babies born with this ailment.  The random variable represents     the number of babies that recover fully.  Decide whether the     experiment is a binomial experiment.  If it is, identify a success,     specify the values of n, p, and q, and list the possible values of the     random variable x.

Is the experiment a binomial experiment?

No

Yes

What is a success in this experiment?

Baby recovers

Baby doesn’t     recover

This is not a     binomial experiment

Specify the value of n.  Select the correct choice below and fill in any answer boxes in your choice.

N = 6

This is not a     binomial experiment

Specify the     value of p. Select the correct choice below and fill in any answer boxes     in your choice.

p = 0.7

This is not a     binomial experiment

Specify the     value of q. Select the correct choice below and fill in any answer boxes     in your choice.

q = 0.3

This is not a     binomial experiment

List the possible values of the random variable x.

X = 0, 1, 2,…,5

X = 0, 1, 2,…6

X = 1, 2, 3,…6

This is not a     binomial experiment

    Find the mean,     variance, and standard deviation of the binomial distribution with the     given values of n and p.

N = 129, p = 0.43

The mean, µ is 55.5 (Round to the nearest tenth as needed.)

The variance, is 31.6 (Round to the nearest tenth as needed.)

The standard deviation, is 5.6 (Round to the nearest tenth as needed.)

  Find the mean,     variance, and standard deviation of the binomial distribution with the     given values of n and p.

N = 121, p = 0.27

The mean, µ is 32.7 (Round to the nearest tenth as needed.)

The variance, is 23.8 (Round to the nearest tenth as needed.)

The standard deviation, is 4.9 (Round to the nearest tenth as needed.)

  48% of men     consider themselves professional baseball fans. You randomly select 10 men     and ask each if he considers himself a professional baseball fan.      Find the probability that the number who consider themselves baseball fans     is (a) exactly eight, (b) at least eight, and (c) less than eight.      If convenient, use technology to find the probabilities.

P(8) = .034     (Round to the nearest thousandth as needed)

P(x≥8) =     042(Round to the nearest thousandth as needed)

P(x<8) =     958(Round to the nearest thousandth as needed)

   Seventy-five     percent of households say they would feel secure if they had $50,000 in     savings. You randomly select 8 households and ask them if they would feel     secure if they had $50,000 in savings.  Find the probability that the     number that say they would feel secure is (a) exactly five, (b) more than     five, and (c) at most five.

Find the     probability that the number that say they would feel secure is exactly     five.

P(5) = .208

(Round to three decimal places as needed)

Find the     probability that the number sat they would feel secure is more than five.

P(x>5) = .678

(Round to three decimal places as needed)

Find the     probability that the number that say they would feel secure is at most     five.

P(x≤5) = .322

(Round to three decimal places as needed)

   Sixty-five     percent of households say they would feel secure if they had $50,000 in     savings. You randomly select 8 households and ask them if they would feel     secure if they had $50,000 in savings.  Find the probability that the     number that say they would feel secure is (a) exactly five, (b) more than     five, and (c) at most five.

Find the     probability that the number that say they would feel secure is exactly     five.

P(5) = 0.279

(Round to three decimal places as needed)

Find the     probability that the number sat they would feel secure is more than five.

P(x>5) = 0.428

(Round to three decimal places as needed)

Find the     probability that the number that say they would feel secure is at most     five.

P(x≤5) = 0.572

(Round to three decimal places as needed)

   34% of adults     say cashews are their favorite kind of nut. You randomly select 12 adults     and ask each to name his or her favorite nut.  Find the probability     that the number who say cashews are their favorite nut is (a) exactly     three, (b) at least four, and (c) at most two. If convenient, use technology     to find the probabilities.

P(3) = .205     (Round to the nearest thousandth as needed.)

P(x > 4)     = 626 (Round to the nearest thousandth as needed.)

P(x < 2)     =.169 (Round to the nearest thousandth as needed)

  33% of adults     say cashews are their favorite kind of nut. You randomly select 12 adults     and ask each to name his or her favorite nut.  Find the probability     that the number who say cashews are their favorite nut is (a) exactly     three, (b) at least four, and (c) at most two. If convenient, use     technology to find the probabilities.

P(3) = 215     (Round to the nearest thousandth as needed.)

P(x > 4)     = 597 (Round to the nearest thousandth as needed.)

P(x < 2)     =188 (Round to the nearest thousandth as needed)

  21% of college     students say they use credit cards because of the rewards program. You     randomly select 10 college students and ask each to name the reason he or     she uses credit cards. Find the probability that the # of college students     who say they use credit cards because of the rewards program is (a)     exactly 2, (b) more than 2, and (c) between 2 and 5 inclusive. If     convenient, use technology to find the probabilities.

P(2) = .205     (Round to the nearest thousandth as needed.)

P(X >2)     = 626 (Round to the nearest thousandth as needed.)

P(X <5)     = .169 (Round to the nearest thousandth as needed.)

  38% of college     students say they use credit cards because of the rewards program. You     randomly select 10 college students and ask each to name the reason he or     she uses credit cards. Find the probability that the # of college students     who say they use credit cards because of the rewards program is (a)     exactly 2, (b) more than 2, and (c) between 2 and 5 inclusive. If     convenient, use technology to find the probabilities.

P(2) = 142(Round     to the nearest thousandth as needed.)

P(X >2)     = 798(Round to the nearest thousandth as needed.)

P(X <5)     = 805 (Round to the nearest thousandth as needed.)

  36% of women     consider themselves fan of professional baseball. You randomly select 6     women and ask each if they consider themselves a fan of professional     baseball.

Construct a     binomial distribution using n = 6 and p = 0.36

X                     P(x)

.069

.232

.326

.245

.103

.023

.002

  Choose the     correct histogram for this distribution below.

  Describe the     shape of the histogram

Skewed right

Skewed left

Symmetrical

None of these

Find the mean of     the binomial distribution

µ = 2.2 (round to the nearest 10th as needed)

( e ) find the variance of the binomial distribution.

= 1.4 (round to the nearest 10th as needed.)

( f ) Find the standard deviation of the binomial distribution.

= 1.2 (round to the nearest 10th as needed)

( g ) Interpret the results in the context of the real-life situation. What values of the random variable would you consider unusual? Explain your reasoning.

On average, 2.2 out of 6 women consider themselves baseball fans, with a standard deviation of 1.2 women. The values x=6 and x=5 would be unusual because their probabilities are less than 0.05.

  38% of women     consider themselves fan of professional baseball. You randomly select 6     women and ask each if they consider themselves a fan of professional     baseball.

Construct a     binomial distribution using n = 6 and p = 0.38

X                     P(x

057

209

320

262

120

029

6                    0.003

  Choose the     correct histogram for this distribution below.

 C Describe the shape of the histogram

Skewed right

Skewed left

Symmetrical

None of these

Find the mean of     the binomial distribution

µ = 2.3 (round to the nearest 10th as needed)

( e ) find the variance of the binomial distribution.

= 1.4 (round to the nearest 10th as needed.)

( f ) Find the standard deviation of the binomial distribution.

= 1.2 (round to the nearest 10th as needed)

( g ) Interpret the results in the context of the real-life situation. What values of the random variable would you consider unusual? Explain your reasoning.

On average, 2.3 out of 6 women consider themselves baseball fans, with a standard deviation of 1.2 women. The values x=6 and x=5 would be unusual because their probabilities are less than 0.05.

   Given that x has     a Poisson distribution with µ = 3, what is the probability that x = 5?

P(5) ≈ 0.1008 (round to 4 decimal places as needed.)

  Given that x has     a Poisson distribution with µ = 4, what is the probability that x = 3?

P(3) ≈ 0.1954 (round to 4 decimal places as needed.)

   Given that x has     a Poisson distribution with µ = 1.6, what is the probability that x = 5?

P(5) ≈ 0.176 (round to 4 decimal places as needed.)

  Given that x has     a Poisson distribution with µ =0.5, what is the probability that x = 0?

P(0) ≈ 0.6065 (round to 4 decimal places as needed.)

  Decide which     probability distribution – binomial, geometric, or Poisson – applies to     the question. You do not need 2 answer the question.

Given: of students ages 16 to 18 with A or B averages who plan to attend college after graduation, 60% cheated to get higher grades. 10 randomly chosen students with A or B to attend college after graduation were asked if they cheated to get higher grades. Question: what is the probability that exactly two students answered no?

Poisson     distribution

Binomial     distribution

Geometric     distribution

  Decide which     probability distribution – binomial, geometric, or Poisson – applies to     the question. You do not need 2 answer the question. Instead, justify your     choice.     Question: what is the probability that 2 many tankers will arrive on a     given day?

You are     interested in counting the number of successes out of n trials.

You are     interested in counting the number of occurrences that take place within a     given unit of time.

You are interested     in counting the number of trials until the first success.

  Decide which     probability distribution – binomial, geometric, or Poisson – applies to     the question. You do not need to answer the question.

Given: Of students ages 16 to 18 with A or B averages who plan to attend college after graduation, 65% cheated to get higher grades.  Ten randomly chosen students with A or B attend college after graduation were asked if the cheated to get higher grades.  Question: What is the probability that exactly two students answered no?

What type of distribution applies to the given question?

  Binomial     distribution

Geometric     distribution

Poisson     distribution

  Decide which     probability distribution – binomial, geometric, or Poisson – applies to     the question. You do not need to answer the question.  Instead,     justify your choice.

Given: The mean number of oil tankers at a port city is 12 per day.  The port has facilities to handle up to 18 oil tankers in a day.

Choose the correct probability distribution below.

You are interested     in counting the number of occurrences that take place within a given unit     of time.

You are     interested in counting the number of successes out of n trials.

You are     interested in counting the number of trials until the first success.

  Find the indicated     probabilities using the geometric distribution or Poisson distribution.     Then determine if the events are unusual. If convenient, use a Poisson     probability table or technology to find the probabilities.

Assume the probability that you will make a sale on any given telephone call is 0.14. Find the probability that you (a) make your first sale on the fifth call, (b) make your sale on the 1st, 2nd, or 3rd call, and (c) do not make a sale on the first 3 calls. (a) P(make your first sale on the fifth call) = 0.077 (Round to three decimal places as needed.)

b) P(make your     sale on the first, second, or third call) = 364

(Round to three decimal places as needed.)

c) P(do not make     a sale on the first three calls) = 636

(Round to three decimal places as needed.)

 Which of the events are unusual?  Select all that apply.

The event in     part (a), “make your first sale on the fifth call”, is unusual

The event in     part (b), “make you sale on the first, second, or third call”, is unusual

The event in     part ©, “do not make a sale on the first three calls”, is unusual

None of the     events are unusual

  Find the     indicted probabilities using the geometric distribution or Poisson     distribution. Then determine if the events are unusual. If convenient, use     a Poisson probability table or technology to find the probabilities.     A newspaper finds the mean number of typographical errors per page is     four. Find the probability that (a) exactly five typographical errors are     found on a page, (b) at most five typographical errors are found on a     page, and (c) more than five typo errors are found on a page.     (a) P(exactly five typo errors are found on a page) = 1563     (Round to four decimal places as needed.)

P(at most five     typographical errors are found on a page) = 7851

(Round to four decimal places as needed.)

P(more than     fivetypo errors are found on a page) = 2149

(Round to four decimal places as needed)

 Which of the events are unusual?  Select all that apply.

The event in     part (a) is unusual.

The event in     part (b) is unusual.

The event in part     (c) is unusual.

None of the     events are unusual

   Find the     indicted probabilities using the geometric distribution or Poisson     distribution. Then determine if the events are unusual. If convenient, use     a Poisson probability table or technology to find the probabilities.

A major hurricane is a hurricane with winds of 111 mph or greater. During the lsat century, the mean # of major hurricanes to strike a certain country’s mainland per year was about 0.46. Find the probability that in a given year (a) exactly one major hurricane will strike the mainland, (b) at most one major hurricane will strike the mainland, and (c) more than one major hurricane will strike the mainland. (a) P(exactly one major hurricane will strike the mainland) = 0.290 (Round to three decimal places as needed.)

b) P(at most one     major hurricane will strike the mainland) = 921

(Round to three decimal places as needed.)

© P(more than one major hurricane will strike the mainland) = 0.079

(Round to three decimal places as needed.)

 Which of the events are unusual?  Select all that apply.

The event in     part (a) is unusual.

The event in     part (b) is unusual.

The event in     part (c) is unusual.

None of the events are unusual  

   ops:lis=����)P

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 4 DQ Discrete Probability Variables - latest

DeVry MATH 221 Week 4 DQ Discrete Probability Variables - latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3357&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Week 4 DQ Discrete Probability Variables

What are examples of variables that follow a binomial probability distribution? What are examples of variables that follow a Poisson distribution? When might you use a geometric probability?

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 3 quiz with Formulas - Latest

DeVry MATH 221 Week 3 quiz with Formulas - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3352&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Week 3 Quiz - 2015

Use the Venn diagram to identify the population and the sample.

Choose the correct description of the population.

The number of home owners in the state

The income of home owners in the state who own a car

The income of home owners in the state

The number of home owners in the state who own a car

Choose the correct description of the sample

The income of home owners in the state who own a car

The income of home owners in the state

The number of home owners in the state who own a car

The number of home owners in the state

Hint: Population is the entire group about which information is needed.

A sample is a smaller group of members of a population selected to represent the population.

Determine whether the variable is qualitative or quantitative.

Favorite sport

Is the variable qualitative or quantitative?

Qualitative

Quantitative

Hint: Since the favorite sport cannot be expressed in numerical terms, it is qualitative.

Students in an experimental psychology class did research on depression as a sign of stress. A test was administered to a sample of 30 students. The scores are shown below.

43  50  10  91  76  35  64  36  42  72  53  62  35  74  50

72  36  28  38  61  48  63  35  41  22  36  50  46  85  13

To find the 10% trimmed mean of a data set, order the data, delete the lowest 10% of the entries and highest 10% of the entries, and find the mean of the remaining entries.  Complete parts (a) through (c).

Find the 10% trimmed mean for the data.

The 10% trimmed mean is 48.8.(Round to the nearest tenth as needed.)

Hint: Please see the excel spreadsheet.

Compare the four measures of central tendency, including the midrange.

Mean = 48.9 (Round to the nearest tenth as needed.)

Median = 47

Mode = 35, 36, 50 (Use a comma to separate answers as needed.)

The midrange is found by the following formula.

(Maximum data entry) + (Minimum data entry)

2

 Midrange = 50.5 (Round to the nearest tenth as needed.)

Hint:

Mean =  = 48.9

Median = item when the items are arranged in ascending or descending order of magnitude.

= item = 15.5th item

=

= (46 + 48)/2

= 47

Mode = most frequently occurring item

Here the observations 35, 36 and 50 occur 3 times and hence all these are modes.

Midrange = (Maximum data entry) + (Minimum data entry)

2

= (10 + 91)/2 = 50.5

 © What is the benefit of using a trimmed mean versus using a mean found using all data entries?

It simply decreases the number of computations in finding the mean.

It permits the comparison of the measures of central tendency.

It permits finding the mean of a data set more exactly.

It eliminates potential outliers that could affect the mean of the entries.

 Construct a frequency distribution for the given data set using 6 classes. In the table, include the midpoints, relative frequencies, and cumulative frequencies. Which class has the greatest frequency and which has the least frequency?

Amount (in dollars) spent on books for a semester

457  146  287  535  442  543  46  405  496  385  517  56  33  132  64

99  378  145  30  419  336  228  376  227  262  340  172  116  285

Complete the table, starting with the lowest class limit.  Use the minimum data entry as the lower limit of the first class. (Type integers or decimals rounded to the nearest thousandth as needed.)

Which class has the greatest frequency?

The class with the greatest frequency is from 374 to 459.

Which class has the least frequency?

The class with the least frequency is from 288 to 373.

Hint:

Minimum value = 30

Maximum value = 543

Number of classes required = 6

Now (543 – 30)/6 = 85.5

Hence the upper limit of the first class is 30 + 85 = 115

2nd class lower limit = 116

2nd class upper limit = 116 + 85 = 201

3rd class lower limit = 202

3rd class upper limit = 202 + 85 = 287

4th class lower limit = 288

4th class upper limit = 288 + 85 = 373

5th class lower limit = 374

5th class upper limit = 374 + 85 = 459

6th class lower limit = 460

6th class upper limit = 460 + 85 = 545

For the calculations of frequency, midpoint, etc. please see the excel spreadsheet.

Identify the data set’s level of measurement.

The nationalities listed in a recent survey (for example, American, German, or Brazilian)

Nominal

Ordinal

Interval

Ratio

Hint: Since the nationalities do not have any numerical value or order, it is nominal.

Explain the relationship between variance and standard deviation. Can either of these measures be negative?

Choose the correct answer below.

The standard deviation is the negative square root of the variance. The standard deviation can be negative but the variance can never be negative.

The standard deviation is the positive square root of the variance. The standard deviation and variance can never be negative.  Squared deviations can never be negative.

The variance is the negative square root of the standard deviation. The variance can be negative but the standard deviation can never be negative.

The variance is the positive square root of the standard deviation. The standard deviation and variance can never be negative.  Squared deviations can never be negative.

Hint:

Variance =

Standard deviation =

That is, standard deviation is the positive square root of variance.

For the following data (a) display the data in a scatter plot, (b) calculate the correlation coefficient r, and (c) make a conclusion about the type of correlation.

The number of hours 6 students watched television during the weekend and the scores of each student who took a test the following Monday.

Hours spent watching TV, x   0          1          2          3          3          5

Test score, y                            98        90        84        74        93        65

Choose the correct scatter plot below.

The correlation coefficient r is -0.844 (Round to three decimal places as needed)

Which of the following best describes the type of correlation that exists between number of hours spent watching television and test scores?

Strong negative linear correlation

No linear correlation

Weak negative linear correlation

Strong positive linear correlation

Weak positive linear correlation

Hint:

For scatter plot and calculation of correlation coefficient, please see the excel spreadsheet.

Since the absolute value of correlation coefficient close to 1, it is strong. Since the correlation coefficient is negative, there is negative correlation. Hence there is a strong negative linear correlation exists between the variables.

Suppose a survey of 526 women in the United States found that more than 70% are the primary investor in their household. Which part of the survey represents the descriptive branch of statistics?

Choose the best statement of the descriptive statistic in the problem.

There is an association between the 526 women and being the primary investor in their household.

526 women were surveyed.

70% of women in the sample are the primary investor in their household.

There is an association between U.S. women and being the primary investor in their household.

Choose the best inference from the given information.

There is an association between the 526 women and being the primary investor in their household.

There is an association between U.S. women and being the primary investor in their household.

70% of women in the sample are the primary investor in their household

526 women were surveyed.

Hint:

Since 70% of women in the sample are the primary investor in their household, we can infer that there is an association between U.S. women and being the primary investor in their household.

Identify the sampling technique used.

A community college student interviews everyone in a particular statistics class to determine the percentage of students that own a car.

Radom

Cluster

Convenience

Stratified

Systematic

Hint: A convenience sample is made up of people who are easy to reach. Here the community college student interviews everyone in a particular statistics class which is easy for him. Hence it is a convenient sampling.

Use the frequency polygon to identify the class with the greatest, and the class with the least frequency.

What are the boundaries of the class with the greatest frequency?

5-30.5

25-31

5-29.5

28-31

What are the boundaries of the class with the least frequency?

10-13

5-11.5

7-13

5-12.5

Hint:

Clearly the greatest frequency is 14 and is in the boundary 26.5-29.5.

Least frequency is 1 and is in the boundary 8.5-11.5.

Determine whether the given value is a statistic or a parameter

In a study of all 2377 students at a college, it is found that 35% own a computer

Choose the correct statement below.

Parameter because the value is a numerical measurement describing a characteristic of a population.

Statistic because the value is a numerical measurement describing a characteristic of a population.

Statistic because the value is a numerical measurement describing a characteristic of a sample.

Parameter because the value is a numerical measurement describing a characteristic of a sample.

 Compare the three data sets

Which data set has the greatest sample standard deviation?

Data set (iii), because it has more entries that are farther away from the mean

Data set (ii), because it has two entries that are far away from the mean.

Data set (i), because it has more entries that are close to the mean.

Which data set has the least sample standard deviation?

Data set (i), because it has more entries that are close to the mean.

Data set (ii), because it has less entries that are farther away from the mean.

Data set (iii), because it has more entries that are farther away from the mean.

How are the data sets the same? How do they differ?

The three data sets have the same standard deviations but have different means.

The three data sets have the same mean, median and mode but have different standard deviation.

The three data sets have the same mean and mode but have different medians and standard deviations.

The three data sets have the same mode but have different standard deviations and means

 Decide which method of data collection you would use to collect data for the study.

A study of the effect on the human digestive system of a popular soda made with a caffeine substitute.

Choose the correct answer below.

Observational Study

Simulation

Survey

Experiment

 Use the given frequency distribution to find the:

Class width

Class midpoint of the first class

Class boundaries of the first class

(a) 4 (b) 137.5 (c) 134.5-139.5

(a) 5 (b) 137 (c) 135-139

(a) 5 (b) 137 (c) 134.5-139.5

(a) 4 (b) 137.5 (c) 135-139

Hint:

Class width = 139.5 – 134.5 = 5

Class midpoint of the first class = (135 + 139)/2 = 137

Class boundaries of the first class = 134.5-139.5

Consider the following sample data values.

5          14        15        21        16        13        9          19

(a) Calculate the range

(b) Calculate the variance

© Calculate the standard deviation

 The range is 16. (Type an integer or a decimal)

The sample variance is 57. (Type an integer or decimal rounded to two decimal places as needed)

The sample standard deviation is 15. (Type an integer or decimal rounded to two decimal places as needed)

Hint:

X

X - X̅

(X - X̅)^2

5

-9

81

14

00

15

1

1

21

7

49

16

2

4

13

-1

1

9

-5

25

19

5

25

112

 186

 Range = Highest observed value – Lowest observed value

 = 21 – 5

 = 16

Variance =  = 26.57

Standard deviation = = 5.15

Find the equation of the regression line for the given data. Then construct a scatter plot of the data and draw the regression line.  (the pair of variables have a significant correlation.)  Then use the regression equation to predict the value of y for each of the given x-values, if meaningful.  The number of hours 6 students spent for a test and their scores on that test are shown below.

Find the regression equation.

^

Y = 6.643 x + (31.238) (Round to three decimal places as needed)

 Choose the correct graph below.

  Predict the value of y for x = 4. Choose the correct answer below.

1

8

8

Not meaningful

Predict the value of y for x = 4.5. Choose the correct answer below.

1

8

0Not meaningful

© Predict the value of y for x = 12.  Choose the correct answer below.

8

1

0Not meaningful

 Predict the value of y for x = 2.5. Choose the correct answer below.

57.8

47.8

111.0

Not meaningful

 Hint:

Scatter plot is given by,

Regression equation is given by,

Y = 6.643 x + 31.238 (Please see the excel sheet for workings)

(a) When x = 4,

Y = (6.643 * 4) + 31.238 = 57.8

(b) When x = 4.5,

Y = (6.643 * 4.5) + 31.238 = 61.1

(c) When x = 12,

Clearly x = 12 is outside the range of values of independent variable. Hence not meaningful.

(d) When x = 2.5,

Y = (6.643 * 2.5) + 31.238 = 47.8

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 3 Homework - Latest

DeVry MATH 221 Week 3 Homework - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3316&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

 MATH 221 Homework Week 3

The access code     for a car’s security system consist of four digits. The first digit cannot     be zero and the last digit must be odd.  How many different codes are     available?

The number of different codes available is 4500.

  The access code     for a car’s security system consist of four digits. The first digit cannot     be 6 and the last digit must be even or zero.  How many different     codes are available?

The number of different codes available is 4500.

A probability     experiment consists of rolling a 6-sided die. Find the probability of the     event below:

Rolling a number is less than 5

The probability is 0.667.

A probability     experiment consists of rolling a 6-sided die. Find the probability of the     event below:

Rolling a number is less than 4

The probability is .5.

Use the     frequency distribution, which shows the responses of a survey of college     students when asked, “How often do you wear a seat belt when riding in a     car driven by someone else?” Find the following probabilities of responses     of college students from the survey chosen at random.

Response                     Probability

Never                          0.021               (Round to the nearest thousandth as needed)

Rarely                          0.062               (Round to the nearest thousandth as needed)

Sometimes                   0.110               (Round to the nearest thousandth as needed)

Most of the time         .272                 (Round to the nearest thousandth as needed)

Always                        .536     (Round to the nearest thousandth as needed)

  Use the     frequency distribution, which shows the responses of a survey of college     students when asked, “How often do you wear a seat belt when riding in a     car driven by someone else?” Find the following probabilities of responses     of college students from the survey chosen at random.

 Response                     Probability

Never                          0.031               (Round to the nearest thousandth as needed)

Rarely                          0.066               (Round to the nearest thousandth as needed)

Sometimes                   0.111               (Round to the nearest thousandth as needed)

Most of the time         .288                 (Round to the nearest thousandth as needed)

Always                        .503     (Round to the nearest thousandth as needed)

Determine     whether the events E and F are independent or dependent. Justify your     answer.

E: A person     having an at-fault accident.

F: The same person being prone to road rage.

E and F are     dependent because having an at-fault accident has no effect on the     probability of a person being prone to road rage.

E and F are     dependent because being prone to road rage can affect the probability of a     person having an at-fault accident.

E and F are     independent because having an at-fault accident has no effect on the     probability of a person being prone to road rage.

E and F are     independent because being prone to road rage has no effect on the     probability of a person having an at-fault accident.

  E: A randomly     selected person accidentally killing a spider.

F: Another randomly selected person accidentally swallowing a spider.

E can affect the     probability of F, even if the two people are randomly selected, so the     events are dependent.

E can affect the     probability of F because the people were randomly selected, so the events     are dependent.

E cannot affect     F and vice versa because the people were randomly selected, so the events     are independent.

E cannot affect     F because “person 1 accidently killing a spider” could never occur, so the     events are neither dependent nor independent.

E: The consumer     demand for synthetic diamonds.

F: The amount of research funding for diamond synthesis.

The consumer     demand for synthetic diamonds could not affect the amount of research     funding for diamond synthesis, so E and F are independent.

The consumer     demand for synthetic diamonds could affect the amount of research funding     for diamond synthesis, so E and F are dependent.

The amount of     research funding for diamond synthesis could affect the consumer demand     for synthetic diamonds, so E and F are dependent.

 C E: The unusually foggy weather in London on May 8

F: The number of car accidents in London on May 8

  The unusually     foggy weather in London on May 8 could not affect the number of car     accidents in London on May 8, so E and F are independent.

The number of     car accidents in London on May 8 could affect the unusually foggy weather     in London on May 8, so E and F are dependent

The unusually     foggy weather in London on May 8 could affect the number of car accidents     in London on May 8, so E and F are dependent

The table below     shows the results of a survey in which 147 families were asked if they own     a computer and if they will be taking a summer vacation this year.

Summer Vacation This Year

                                                Yes                   No         Total

Own  a              Yes                   46                     11                     57

Computer         No                    56                     34                     90

Total                                          102                   45                     147

Find the     probability that a randomly selected family is not taking a summer     vacation this year.

The probability is .306 (Round to the nearest thousandth as needed.)

Find the     probability that a randomly selected family owns a computer.

The probability is .388 (Round to the nearest thousandth as needed.)

Find the     probability that a randomly selected family is taking a summer vacation     this year given that they own a computer.

The probability is .807 (Round to the nearest thousandth as needed.)

Find the probability     that a randomly selected family is taking a summer vacation this year and     owns a computer.

The probability is .313 (Round to the nearest thousandth as needed.)

Are the events     of owning a computer and taking a summer vacation this year independent or     dependent events?

Dependent

  The table below     shows the results of a survey in which 147 families were asked if they own     a computer and if they will be taking a summer vacation this year.

Summer Vacation This Year

Yes                   No                    Total

Own  a              Yes                   47                     11                     58

Computer         No                    56                     33                     89

Total                                          103                   44                     147

Find the     probability that a randomly selected family is not taking a summer     vacation this year.

The probability is .299 (Round to the nearest thousandth as needed.)

Find the     probability that a randomly selected family owns a computer.

The probability is .395 (Round to the nearest thousandth as needed.)

Find the     probability that a randomly selected family is taking a summer vacation     this year given that they own a computer.

The probability is .810 (Round to the nearest thousandth as needed.)

Find the     probability that a randomly selected family is taking a summer vacation     this year and owns a computer.

The probability is .320 (Round to the nearest thousandth as needed.)

Are the events     of owning a computer and taking a summer vacation this year independent or     dependent events?

Dependent

A distribution     center receives shipments of a product from three different factories in     the quantities of 50, 30, and 20. Three times a product is selected at     random, each time without replacement.  Find the probability that (a)     all three products came from the second factory and (b) none of the three     products came from the second factory.

The probability     that all three products came from the second factory is 007

(Round to the nearest thousandth as needed.)

The probability     that none of the three products came from the second factory is .508

(Round to the nearest thousandth as needed.)

   A standard deck     of cards contains 52 cards. One card is selected from the deck.

Compute the     probability of randomly selecting a spade or heart

Compute the     probability of randomly selecting a spade or heart or diamond

Compute the     probability of randomly selecting a seven or club

  P(spade of     heart)=.5 (Type an integer or a simplified fraction.)

P(spade or heart     or diamond)=.75 (Type an integer or a simplified fraction.)

P(seven or     club)=4 (Type an integer or a simplified fraction.)

13

   A standard deck     of cards contains 52 cards. One card is selected from the deck.

Compute the     probability of randomly selecting a three or eight

Compute the     probability of randomly selecting a three or eight of king

Compute the     probability of randomly selecting a queen or diamond

  P(spade of     heart)=2 (Type an integer or a simplified fraction.)

13

P(spade or heart     or diamond)=3 (Type an integer or a simplified fraction.)

13

P(seven or     club)=4 (Type an integer or a simplified fraction.)

13

  The percent     distribution of live multiple-delivery births (three or more babies) in a     particular year for a women 15 to 54 years old is shown in the pie chart.     Find each probability.

Randomly     selecting a mother 30 – 39 years old

P(30 to 39) ≈ .621 (Round to the nearest thousandth as needed.)

Randomly     selecting a mother not 30-39 years old

P(not 30 to 39) ≈ .379 (Round to the nearest thousandth as needed.)

C, Randomly selecting a mother less than 45 years old

P(less than 45) ≈ .967 (Round to the nearest thousandth as needed.)

Randomly     selecting a mother at least 20 years old

P(at least 20) ≈ .985 (Round to the nearest thousandth as needed.)

   The percent     distribution of live multiple-delivery births (three or more babies) in a     particular year for a women 15 to 54 years old is shown in the pie chart.     Find each probability.

   Randomly     selecting a mother 30 – 39 years old

P(30 to 39) ≈ .619 (Round to the nearest thousandth as needed.)

Randomly     selecting a mother not 30-39 years old

P(not 30 to 39) ≈ .381 (Round to the nearest thousandth as needed.)

C, Randomly selecting a mother less than 45 years old

P(less than 45) ≈ .968 (Round to the nearest thousandth as needed.)

Randomly     selecting a mother at least 20 years old

P(at least 20) ≈ .984 (Round to the nearest thousandth as needed.)

  The table below     shows the number of male and female students enrolled in nursing at a     university for a certain semester. A student is selected at random.      Complete parts (a) through (d).

Nursing majors                        Non-nursing majors                 Total

Males                                       92                                1019                            1111

Females                                   700                              1725                            2425

Total                                        792                              2744                            3536

Find the     probability that the student is male or a nursing major

(Round to the nearest thousandth as needed.)

 P (being male or being nursing major) = .512

Find the     probability that the student is female or not a nursing major.

(Round to the nearest thousandth as needed.)

P( being female or not a nursing major) = .974

Find the     probability that the student is not female or a nursing major

(Round to the nearest thousandth as needed.)

P(not being female or being a nursing major) = .512

Are the events     “being male” and “being a nursing major” mutually exclusive?

No, because     there are 92 males majoring in nursing

No, because one     can’t be male and a nursing major at the same time

Yes, because one     can’t be male and a nursing major at the same time

Yes, because there are 97 males majoring in nursing

  The table below     shows the number of male and female students enrolled in nursing at a     university for a certain semester. A student is selected at random.      Complete parts (a) through (d).

Nursing majors                        Non-nursing majors                 Total

Males                                       97                                1017                            1114

Females                                   700                              1727                            2427

Total                                        797                              2744                            3541

Find the     probability that the student is male or a nursing major

(Round to the nearest thousandth as needed.)

 P (being male or being nursing major) = .512

Find the     probability that the student is female or not a nursing major.

(Round to the nearest thousandth as needed.)

P( being female or not a nursing major) = .972

Find the     probability that the student is not female or a nursing major

(Round to the nearest thousandth as needed.)

P(not being female or being a nursing major) = .512

Are the events     “being male” and “being a nursing major” mutually exclusive?

No, because     there are 97 males majoring in nursing

No, because one     can’t be male and a nursing major at the same time

Yes, because one     can’t be male and a nursing major at the same time

Yes, because     there are 97 males majoring in nursing

  Outside a home,     there is an10-key keypad with letters A, B, C, D, E, F, G and H that can     be used to open the garage if the correct ten-letter code is entered. Each     key may be used only once.  How many codes are possible?

The number of possible codes is 3628800.

  How many     different 10-letter words (real or imaginary) can be formed from the     following letters?

Z, V, U, G, X, V, H, G, D

302400 ten-letter words (real or imaginary) can be formed with the given letters.

   How many     different 10-letter words (real or imaginary) can be formed from the     following letters?

K, I, B, W, E, Z, I, O, R, Z

907200 ten-letter words (real or imaginary) can be formed with the given letters.

   A horse race has     13 entries and one person owns 2 of those horses. Assuming that there are     no ties, what is the probability that those four horses finish first,     second, third, and fourth (regardless of order)?

The probability that those two horses finish first, second, third, and fourth is 0.0128.

(Round to the nearest thousandth as needed.)

  A horse race has     13 entries and one person owns 4 of those horses. Assuming that there are     no ties, what is the probability that those four horses finish first,     second, third, and fourth (regardless of order)?

The probability that those four horses finish first, second, third, and fourth is .0014.

(Round to the nearest thousandth as needed.)

Determine the     required value of the missing probability to make the distribution a     discrete probability distribution.

X         P(x)

3          0.19

4          ?

5          0.34

6          0.28

P(4) = .19 (Type an integer or a decimal)

  Determine the     required value of the missing probability to make the distribution a     discrete probability distribution.

X         P(x)

3          0.16

4          ?

5          0.38

6          0.17

P(4) = .29 (Type an integer or a decimal)

   A frequency     distribution is shown below. Complete parts (a) through (e).

Dogs                0          1          2          3          4          5

Household       1324    436      162      46        27        15

Use the     frequency distribution to construct a probability distribution.

X         P(x)

0          0.659

1          0.217

2          0.081

3          0.023

4          0.013

5          0.007

(Round to the nearest thousandth as needed.)

Find the mean of     the probability distribution

µ = 0.5 (Round to the nearest thousandth as needed.)

Find the     variance of the probability distribution

= 8 (Round to     the nearest tenth as needed)

Find the     standard deviation of the probability distribution

= 9 (Round to     the nearest tenth as needed)

Interpret the     results in the context of the real-life situation.

A household on     average has 0.5 dog with a standard deviation of 0.9 dog.

A household on     average has 0.5 dog with a standard deviation of 15 dog.

A household on     average has 0.9 dog with a standard deviation of 0.5 dog.

A household on     average has 0.9 dog with a standard deviation of 0.9 dog.

   Students in a     class take a quiz with eight questions. The number x of questions answered     correctly can be approximated by the following probability     distribution.  Complete parts (a) through (e).

 X         0          1          2          3          4          5          6          7          8

P(x)     0.02     0.02     0.06     0.06     0.14     0.24     0.27     0.12     0.07

  Use the     probability distribution to find the mean of the probability distribution

µ = 5.1 (Round to the nearest tenth as needed)

Use the     probability distribution to find the variance of the probability     distribution

= 3.1 (Round to the nearest tenth as needed)

© Use the probability distribution to find the standard deviation of the probability distribution

(Round to the     nearest tenth as needed)

Use the     probability distribution to find the expected value of the probability     distribution

5.1 (Round to the nearest tenth as needed)

  Interpret the     results

The expected     number of questions answered correctly is 5.1 with a standard deviation of     1.8 questions.

The expected     number of questions answered correctly is 1.8 with a standard deviation of     5.1 questions.

The expected     number of questions answered correctly is 5.1 with a standard deviation of     0.02 questions.

The expected number     of questions answered correctly is 3.1 with a standard deviation of 1.8     questions.

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 3 DQ Statistics in the News - Latest

DeVry MATH 221 Week 3 DQ Statistics in the News - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3348&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Week 3 DQ Statistics in the News

Keep your eyes and ears open as you read or listen to the news this week. Find/discover an example of statistics in the news to discuss the following statement that represents one of the objectives of statistics analysis: “Statistics helps us make decisions based on data analysis.” Briefly discuss how the news item or article meets this objective. Cite your references.

   ion=edi���N]`

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 2 iLab - Latest

DeVry MATH 221 Week 2 iLab - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3344&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

Statistics – Lab Week 2

 Name:

Math221

 Statistical Concepts:

Using Minitab

Graphics

Shapes of Distributions

Descriptive Statistics

Empirical Rule

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 2 Homework  - Latest

DeVry MATH 221 Week 2 Homework  - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3312&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

 DeVry MATH 221 Week 2 Homework  - Latest 2015

 MATH 221 Homework Week 2

  Two variables     have a positive linear correlation. Does the dependent variable increase     or decrease as the independent variable increases?

Choose the correct answer below.

The dependent     variable decreases

The dependent     variable increases

  Discuss the     difference between r and p

Choose the correct answers below.

R represents the sample correlation coefficient.

P represents the population correlation coefficient

   The scatter plot     of a paired data set is shown. Determine whether there is a perfect     positive linear correlation, a strong positive linear correlation, a     perfect negative linear correlation, a strong negative linear correlation,     or no linear correlation between the variables.

  Choose the correct answer below.

  no linear     correlation

strong positive     linear correlation

strong negative     linear correlation

perfect negative     linear correlation

perfect positive     linear correlation

 3 The scatter plot of a paired data set is shown.  Determine whether there is a perfect positive linear correlation, a strong positive linear correlation, a perfect negative linear correlation, a strong negative linear correlation, or no linear correlation between the variables.

   no linear     correlation

strong positive     linear correlation

strong negative     linear correlation

perfect negative     linear correlation

perfect positive     linear correlation

   Identify the     explanatory variable and the response variable.

 A golfer wants to determine if the amount of practice every year can be used to predict the amount of improvement in his game.

The explanatory variable is the amount of practice

The response variable is the amount of improvement in his game

  4 Identify the explanatory variable and the response variable.

 A teacher wants to determine if the amount of textbook used by her students can be used to predict the students’ test scores

The explanatory variable is the type of text book

The response variable is the students’ test scores

  Two variables     have a positive linear correlation. Is the slope of the regression line     for the variables positive or negative?

The slope is     positive. As the independent variable increases the dependent variable     also tends to increase

The slope is     negative. As the independent variable increases the dependent variable     tends to decrease

The slope is     negative. As the independent variable increases the dependent variable     tends to increase.

The slope is     positive. As the independent variable increases the dependent variable     tends to decrease.

  Given a set of     data and a corresponding regression line, describe all values of x that     provide meaningful predictions for y.

Prediction     values are meaningful for all x-values that are realistic in the context     of the original data set.

Prediction     values are meaningful for all x-values that are not included in the     original data set.

Prediction     values are meaningful for all x-values in (or close to) the range of the     original data.

  Match this     description with a description below.

The y-value of a data point corresponding to

Choose the correct answer below.

B

correct answer

M

    Match this     description with a description below.

The y-value for a point on the regression line corresponding to

Choose the correct answer below.

  correct answer

B

M

  Match the     description below with its symbol(s).

The mean of the y-values

Select the correct choice below.

B

  correct answer

  M

  Match the     regression equation with the appropriate graph.

Choose the correct answer below.

 C is the correct answer.

   Match the     regression equation with the appropriate graph.

  D is the correct answer

  11

Use the value of     the linear correlation coefficient to calculate the coefficient of     determination. What does this tell you about the explained variation of     the data about the regression line? About the unexplained variation?

R= -0.312

 Calculate the coefficient of determination

.097 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

9.7% of the variation can be explained by the regression line. (Round to three decimal places as needed)

About the unexplained variation?

90.3% of the variation is unexplained and is due to other factors or to sampling error.  (Round to three decimal places as needed)

 11  Use the value of the linear correlation coefficient to calculate the coefficient of determination.  What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?

R= -0.324

 Calculate the coefficient of determination

0.105 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

10.5 % of the variation can be explained by the regression line. (Round to three decimal places as needed)

About the unexplained variation?

89.5 % of the variation is unexplained and is due to other factors or to sampling error.  (Round to three decimal places as needed)

   Use the value of     the linear correlation coefficient to calculate the coefficient of     determination. What does this tell you about the explained variation of the     data about the regression line? About the unexplained variation?

R = 0.481

 Calculate the coefficient of determination

.231 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

2.1 % of the variation can be explained by the regression line. (Round to three decimal places as needed)

76.9 % of the variation is unexplained and is due to other factors or to sampling error.  (Round to three decimal places as needed)

 12 Use the value of the linear correlation coefficient to calculate the coefficient of determination.  What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?

R = 0.224

 Calculate the coefficient of determination

0.050 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

5 % of the variation can be explained by the regression line. (Round to three decimal places as needed)

95 % of the variation is unexplained and is due to other factors or to sampling error.  (Round to three decimal places as needed)

  The equation     used to predict college GPA (range 0-4.0) is is high school GPA (range     0-4.0) and x2 is college board score (range 200-800). Use the multiple     regression equation to predict college GPA for a high school GPA of 3.5     and college board score of 400.

The predicted college GOA for a high school GPA of 3.5 and college board of 400 is 2.9.  (Round to the nearest tenth as needed).

 13 Use the value of the linear correlation coefficient to calculate the coefficient of determination.  What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?

R = 0.909

 Calculate the coefficient of determination

.826 (Round to three decimal places as needed)

What does this tell you about the explained variation of the data about the regression line?

82.6 % of the variation can be explained by the regression line. (Round to three decimal places as needed)

17.4 % of the variation is unexplained and is due to other factors or to sampling error.  (Round to three decimal places as needed)

  The equation     used to predict the total body weight (in pounds) of a female athlete at a     certain school is is the female athlete’s height (in inches) and x2 is the     female athlete’s percent body fat. Use the multiple regression equation to     predict the total body weight for a female athlete who is 64 inches tall     and has 17% body fat.

The predicted total body weight for a female athlete who is 64 inches tall and has 17% body fat is 140.9 pounds.  (Round to the nearest tenth as needed).

   The equation     used to predict college GPA (range 0-4.0) is is high school GPA (range     0-4.0) and X2 is college board score (range 200-800). Use the multiple     regression equation to predict college GPA for a high school GPA of 3.2     and a college board score of 500.

 The predicted college GPA for a high school GPA of 3.2 and college board score of 500 is 2.9. (Round to the nearest tenth as needed).

  The equation     used to predict the total body weight (in pounds) of a female athlete at a     certain school is is the female athlete’s height (in inches) and X2 is the     female athlete’s percent body fat. Use the multiple regression equation to     predict the total body weight for a female athlete who is 67 inches tall     and has 24% body fat.

 The predicted total body weight for a female athlete who is 67 inches tall and has 24% body fat is 137.8 pounds. (Round to the nearest tenth as needed).

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 2 Discussions Regression - Latest

DeVry MATH 221 Week 2 Discussions Regression - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3340&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Week 2 Discussions Regression

Suppose you are given data from a survey showing the IQ of each person interviewed and the IQ of his or her mother. That is all the information that you have. Your boss has asked you to put together a report showing the relationship between these two variables. What could you present and why?

   zֆ�; `

[fantasticblazedinosaur-blog]

DeVry MATH 221 Week 1 DQ Descriptive Statistics - latest

DeVry MATH 221 Week 1 DQ Descriptive Statistics - latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3336&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Week 1 DQ Descriptive Statistics

If you were given a large data set such as the sales over the last year of our top 1,000 customers, what might you be able to do with this data? What might be the benefits of describing the data?

   _̆�!5`9

[fantasticblazedinosaur-blog]

DeVry MATH 221 Quiz Week 5 - Latest

DeVry MATH 221 Quiz Week 5 - Latest

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3365&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

MATH 221 Quiz Week 5 DeVry

  Sixty percent of     households say they would feel secure if they had $50,000 in savings. You     randomly select 8 households and ask them if they would feel secure if     they had $50,000 in savings.  Find the probability that the number     that say they would feel secure is (a) exactly five, (b) more than five,     and (c) at most five.

Find the     probability that the number that say they would feel secure is exactly     five.

P(5) = 0.279 (Round to three decimal places as needed)

Find the     probability that the number that say they would feel secure is more than     five.

P(x>5) = 0.315 (Round to three decimal places as needed)

Find the     probability that the number that say they would feel secure is at most     five.

P(x≤5) = 0.685 (Round to three decimal places as needed)

 Hint:

 Let X be the number of households say they would feel secure if they had $50,000 in savings. Clearly X is binomial with n = 8 and p = 0.60.  The probability mass function binomial variable is given by .The probability for different values of x are given below.

 X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

00.00065536

0.00065536

00.99934464

1

1

0.00786432

0.00851968

0.00065536

0.99148032

0.99934464

2

0.04128768

0.04980736

0.00851968

0.95019264

0.99148032

3

0.12386304

0.1736704

0.04980736

0.8263296

0.95019264

4

0.2322432

0.4059136

0.1736704

0.5940864

0.8263296

5

0.279

0.685

0.4059136

0.315

0.5940864

6

0.20901888

0.89362432

0.68460544

0.10637568

0.31539456

7

0.08957952

0.98320384

0.89362432

0.01679616

0.10637568

8

0.01679616

1

0.98320384

00.01679616

 (a) P (5) = = 0.279

 (b) P (x > 5) = P (x = 6) + P (x = 7) + P (x = 8)

 = 0.315

 (c) P (x ≤ 5) = 1 – P (x > 5) = 1 – 0.315 = 0.685

   Suppose 80% of     kids who visit a doctor have a fever, and 25% of kids with a fever have     sore throats. What’s the probability that a kid who goes to the doctor has     a fever and a sore throat?

The probability is 0.200. (Round to three decimal places as needed)

 Hint:

 Let A and B respectively be the events that the kids have fever and sore throat.

 Given that P (A) = 80% = 0.80

 P (B|A) = 25% = 0.25

 P (a kid who goes to the doctor has a fever and a sore throat) = P (A∩B)

 = P (B|A) * P (A)

 = 0.25 * 0.80

 = 0.200

  Find the mean,     variance, and standard deviation of the binomial distribution with the     given values of n and p. n = 90, p = 0.8

The mean, µ is 72.0 (Round to the nearest tenth as needed)

The variance, σ2, is 14.4 (Round to the nearest tenth as needed)

The standard deviation, σ is 3.8 (Round to the nearest tenth as needed)

 Hint:

 Mean = np = 90 * 0.8 = 72.0

 Variance, σ2 = np(1 – p) = 90 * 0.8 * (1 – 0.8) = 14.4

 Standard deviation, σ = √14.4 = 3.8

  Use the bar     graph below, which shows the highest level of education received by     employees of a company, to find the probability that the highest level of     education for an employee chosen at random is E.

The probability that the highest level of education for an employee chosen at random is E is 0.069.  (Round to the nearest thousandth as needed)

 Hint:

 P (the highest level of education for an employee chosen at random is E)

 =

 = 6/(6 + 22 + 36 + 16 + 6 + 1)

 = 0.069

  A company that     makes cartons finds that the probability of producing a carton with a     puncture is 0.05, the probability that a carton has a smashed corner is     0.09, and the probability that a carton has a puncture and has a smashed     corner is 0.005. Answer parts (a) and (b) below.

Are the events     “selecting a carton with a puncture” and “selecting a carton with a     smashed corner” mutually exclusive?

No, a carton can     have a puncture and a smashed corner.

Yes, a carton     can have a puncture and a smashed corner

Yes, a carton     cannot have a puncture and a smashed corner

Mo, a carton     cannot have a puncture and a smashed corner

If a quality     inspector randomly selects a carton, find the probability that the carton     has a puncture or has a smashed corner.

The probability that a carton has a puncture or a smashed corner is 0.135.  (Type an integer or a decimal.  Do not round)

  Given that x has     a Poisson distribution with µ = 8, what is the probability that x = 3?

P(3) ≈ 0.0286 (Round to four decimal places as needed)

 Hint:

 The probability density function of Poisson random variable is given by, where λ = 8, x = 3

 P (3) = = 0.0286

  Perform the     indicated calculation.

= 0.0017 (Round to four decimal places as needed)

 Hint:

 5P2 = 5!/(5 – 2)! = 20

 12P4 = 12!/(12 – 4)!  = 11880

 5P2/12P4 = 20/11880 = 0.0017

  A frequency     distribution is shown below. Complete parts (a) through (d)

The number of televisions per household in a small town

Televisions      0          1          2          3

Households     26        448      730      1400

Use the     frequency distribution to construct a probability distribution

X                     P(x)

0                      0.010

1                      0.172

2                      0.280

3                      0.538

(Round to the nearest thousandth as needed)

 Hint: Please see the excel spreadsheet.

  Graph the     probability distribution using a histogram. Choose the correct graph of     the distribution below.

Describe the histogram’s shape.  Choose the correct answer below.

Skewed right

Skewed left

Symmetric

Find the mean of     the probability distribution

µ = 2.3 (round to the nearest tenth as needed)

Find the variance of the probability distribution

σ2 = 0.6 (round to the nearest tenth as needed)

Find the standard deviation of the probability distribution

σ = 0.8 (round to the nearest tenth as needed)

Interpret the     results in the context of the real-life situation.

The mean is 2.3,     so the average household has about 3 television. The standard deviation is     0.6 of the households differ from the mean by no more that about 1     television

The mean is 0.6,     so the average household has about 1 television. The standard deviation is     0.8 of the households differ from the mean by no more that about 1     television

The mean is 2.3,     so the average household has about 2television. The standard deviation is     0.8 of the households differ from the mean by no more that about 1     television

The mean is 0.6,     so the average household has about 1 television. The standard deviation is     2.3 of the households differ from the mean by no more that about 3     television

 Hint:

 X

Households

P (X)

X^2

X * P (X)

X^2 * P (X)

026

0.010

00

01

448

0.172

1

0.17204301

0.172043011

2

730

0.280

4

0.56067588

1.121351767

3

1400

0.538

9

1.61290323

4.838709677

Total

2604

  2.34562212

6.132104455

 μ = ∑ X * P (X) = 2.3

 σ2 = = 6.1321 – (2.3456)^2 = 0.6

 σ = √0.6 = 0.8

  In the general     population, one woman in eight will develop breast cancer. Research has     shown that 1 woman is 650 carries a mutation of the BRCA gene.  Nine     out of 10 women with this mutation develop breast cancer.

Find the     probability that a randomly selected woman will develop breast cancer     given that she has a mutation of the BRCA gene.

The probability that a randomly selected woman will develop breast cancer given that she has a mutation of the BRCA gene is 0.9. (Round to one decimal place as needed)

 Hint:

 P (a randomly selected woman will develop breast cancer given that she has a mutation of the BRCA gene)

 = 9/10

 = 0.9

  Find the     probability that a randomly selected woman will carry the mutation of the     BRCA gene and will develop breast cancer.

The probability that a randomly selected woman will carry the gene nutation and develop breast cancer is 0.0014. (Round to four decimal places as needed)

 Hint:

 P (A∩B) = P (A|B) * P (B)

 = 0.9 * (1/650)

= 0.0014

  Are the events     of carrying this mutation and developing breast cancer independent or     dependent?

Dependent

Independent

 Hint:

 Here P (A|B) ≠ P (A). Hence the events are dependent.

  Students in a     class take a quiz with eight questions. The number x of questions answered     correctly can be approximated by the following probability     distribution.  Complete parts (a) through (e)

X                     0          1          2          3          4          5          6          7          8

P(x)                 0.04     0.04     0.06     0.06     0.12     0.24     0.23     0.14     0.07

Use the     probability distribution to find the mean of the probability distribution

µ= 4.9 (Round to the nearest tenth as needed)

Use the     probability distribution to find the variance of the probability     distribution

σ2=4.0 (Round to the nearest tenth as needed)

Use the     probability distribution to find the standard deviation of the probability     distribution

2.0 (Round to the nearest tenth as needed)

Use the     probability distribution to find the expected value of the probability     distribution

4.9 (Round to the nearest tenth as needed)

Interpret the     results

The expected     number of questions answered correctly is 2.0 with a standard deviation of     4.9 questions.

The expected     number of questions answered correctly is 4 with a standard deviation of     2.0 questions.

The expected     number of questions answered correctly is 4.9 with a standard deviation of     0.04 questions.

The expected     number of questions answered correctly is 4.9 with a standard deviation of     2.0 questions.

 Hint:

 X

P(x)

X^2

X * P (X)

X^2 * P (X)

00.04

00

01

0.04

1

0.04

0.04

2

0.06

4

0.12

0.24

3

0.06

9

0.18

0.54

4

0.12

16

0.48

1.92

5

0.24

25

1.2

6

6

0.23

36

1.38

8.28

7

0.14

49

0.98

6.86

8

0.07

64

0.56

4.48

Total

1

 4.94

28.36

 μ = ∑ X * P (X) = 4.9

 σ2 = = 28.36 – (4.94)^2 = 4.0

 σ = √4 = 2.0

 Expected value = μ = 4.9

   Identify the     sample space of the probability experiment and determine the number of     outcomes in the sample space. Randomly choosing a multiple of 5 between 21     and 49

The sample space is {25,30,35,40,45}  (Use a comma to separate answers as needed.  Use ascending order)

There are 5 outcome(s) in the sample space.

  Decide if the     events shown in the Venn diagram are mutually exclusive.

Are the events mutually exclusive?

Yes

No

 Hint:

 Since the events are overlapping, they have common elements. Hence events are not mutually exclusive.

  Determine     whether the random variable is discrete or continuous.

The number of     free-throw attempts before the first shot is made

The weight of a     T-bone steak

The number of     bald eagles in the country

The number of     points scored during a basketball game

The number of     hits to a website in a day

  Is the number of     free-throw attempts before the first shot is made discrete or continuous?

The random     variable is continuous

The random     variable is discrete

 Hint:

 Since the number of free-throw attempts before the first shot can only be expressed as a whole number, it is discrete.

  Is the weight of     a T-bone steak discrete or continuous?

The random     variable is discrete

The random     variable is continuous

 Hint:

 The weight can take any value and is not restricted to whole numbers. Hence it is continuous.

  Is the number of     bald eagles in the country discrete or continuous?

The random     variable is discrete

The random     variable is continuous

 Hint:

 Since the number of bald eagles in the country can only be expressed as a whole number, it is discrete.

  Is the number of     points scored during a basketball game discrete or continuous?

The random     variable is discrete

The random     variable is continuous

 Hint:

 Since the number of points scored during a basketball game can only be expressed as a whole number, it is discrete.

  Is the number of     hits to a website in a day discrete or continuous?

The random     variable is discrete

The random     variable is continuous

 Hint:

 Since the number of hits to a website in a day can only be expressed as a whole number, it is discrete.

   A survey asks     1100 workers, :Has the economy forced you to reduce the amount of vacation     you plan to take this year?” Fifty-six percent of those surveyed say they     are reducing the amount of vacation.  Twenty workers participating in     the survey are randomly selected.  The random variable represents the     number of workers who are reducing the amount of vacation.  Decide     whether the experiment is a binomial experiment.  If it is, identify     a success, specify the values of n, p, and q, and list the possible values     of the random variable x.

Is the experiment a binomial experiment?

Yes

No

What is a success in this experiment?

Selecting a     worker who is reducing the amount of vacation

Selecting a     worker who is not reducing the amount of vacation

This is not a     binomial experiment

Specify the value of n.  Select the correct choice and fill in any answer boxes in your choice

N=20

This is not a     binomial experiment

Specify the value of p.  Select the correct choice below and fill in any answer boxes in your choice.

P=0.56 (Type an     integer or a decimal)

This is not a     binomial experiment

 Specify the value of q.  Select the correct choice below and fill in any answer boxes in your choice.

Q=0.44(Type an     integer r a decimal)

This is not a     binomial experiment

 List the possible values of the random variable x

X=0, 1, 2,…, 20

X=1, 2, 3,…,     1100

1, 2,…, 20

This is not a     binomial experiment

  Determine     whether the distribution is a discrete probability distribution.

Is the distribution a discrete probability distribution?  Why?  Choose the correct answer below.

Yes, because the     probabilities sum to 1 and are all between 0 and 1, inclusive

No, because the     total probability is not equal to 1

Yes, because the     distribution is symmetric

No, because some     of the probabilities have values greater than 1 or less than 0

  The table below     shows the results of a survey that asked 2872 people whether they are involved     in any type of charity work. A person is selected at random from the     sample.  Complete parts (a) through (e).

Frequency       Occasionally    Not at all         Total

Male                226                  455                  793                  1474

Female             206                  450                  742                  1398

Total                432                  905                  1535                2872

Find the     probability that the person is frequently or occasionally involved in     charity work

P(being frequently involved or being occasionally involved) = 0.466  (Round to the nearest thousandth as needed)

 Hint:

 P(being frequently involved or being occasionally involved) =

 = (432 + 905)/2872

 = 0.466

  Find the     probability that the person is male or frequently involved in charity work

P(being male or being frequently involved) = 0.585

 Hint:

 P(being male or being frequently involved) = P (male) + p (being frequently involved) – P (male and being frequently involved)

 = (1474/2872) + (432/2872) – (226/2872)

 = 0.585

  Find the     probability that the person is female or not involved in charity work at     all

P(being female or not being involved) = 0.763 (Round to the nearest thousandth as needed)

 Hint:

 P(being female or not being involved) = P (female) + P (not being involved) – P (female and not being involved)

 = (1398/2872) + (1535/2872) – (742/2872)

 = 0.763

  Find the     probability that the person is female or not frequently involved in     charity work

P(being female or not being frequently involved) = 0.921 (Round to the nearest thousandth as needed)

 Hint:

 P(being female or not being frequently involved) = P (female) + P (not being frequently involved) – P (being female AND not being frequently involved)

 = (1398/2872) + [(905 + 1535)/2872] – [(450 + 742)/2872)

= 0.921

  Are the events     “being female” and “being frequently involved in charity work” mutually     exclusive?

No, because 206     females are frequently involved in charity work.

Yes, because no     females are frequently involved in charity work.

Yes, because 206     females are frequently involved in charity work.

No, because no     females are frequently involved in charity work.

  For the given     pair of events, classify the two events as independent or dependent.

Swimming all day at the beach

Getting a sunburn

Choose the correct answer below.

The two events     are independent because the occurrence on one does not affect the     probability of the occurrence of the other.

The two events     are dependent because the occurrence of one does not affect the     probability of the occurrence of the other.

The two events     are independent because the occurrence of one affects the probability of     the occurrence of the other.

The two events     are dependent because the occurrence of one affects the probability of the     occurrence of the other.

  Outside a home,     there is a 9-key keypad with letters A, B, C, D, E, F, G, H, and I that     can be used to open the garage if the correct nine-letter code is entered.     Each key may be used only once.  How many codes are possible?

The number of possible codes is 362880.

 Hint:

 Here each letter can be used only once. Therefore, the first letter can be set in 9 different ways using these 9 letters, 2nd letter in 8 different letters and so on.

 Hence the number of possible codes = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

 = 362880

  Determine the     number of outcomes in the event. Decide whether the event is a simple     event or not.

A computer is used to select randomly a number between 1 and 9, inclusive.  Event C is selecting a number less than 5.

 Event C has 4 outcome(s)

Is the event a simple event?

No, because event C has more than one outcome.

  Hint:

 C = {1, 2, 3, 4}

[fantasticblazedinosaur-blog]

DeVry MATH 221 Final Exam - Latest 2015

DeVry MATH 221 Final Exam - Latest 2015

IF You Want To Purcahse A+  Work then Click The Link Below For Instant Down Load

http://www.acehomework.net/wp-admin/post.php?post=3154&action=edit

IF You Face Any Problem Then E Mail Us At  JOHNMATE1122@GMAIL.COM

DeVry MATH 221     Final Exam - Latest

  The table below     shows the number of male and female students enrolled in nursing at a     university for a certain semester. A student is selected at random.     Complete parts (a) through (d).

  Find the     probability that the student is male or a nursing major.

P (being male or being nursing major) = 0.513

Find the     probability that the student is female or not a nursing major.

P(being female or not being a nursing major) = 0.972

Find the     probability that the student is not female or a nursing major

P(not being female or not being a nursing major) = 0.513

Are the events     “being male” and “being a nursing major” mutually exclusive? Explain.

Answer: No, because there are 99 males majoring in nursing.

Hint:

(a) P (male or a nursing major) = P (male) + P (nursing major) – P (male AND a nursing major)

= (1115/3539) + (799/3539) – (99/3539)

= 0.513

(b) P (female or not a nursing major) = P (female) + P (not a nursing major) – P (female AND not a nursing major)

= (2424/3539) + (2740/3539) – (1724/3539)

= 0.972

(c) P (not female or a nursing major) = P (not female) + P (nursing major) – P (not female AND a nursing major)

= (1115/3539) + (799/3539) – (99/3539)

= 0.513

  An employment     information service claims the mean annual pay for full-time male workers     over age 25 without a high school diploma is $22,325. The annual pay for a     random sample of 10 full-time male workers over age 25 without a high     school diploma is listed. At a = 0.10, test the claim that the mean salary     is $22,325. Assume the population is normally distributed.

 20,660 – 21,134 – 22,359 – 21,398 – 22,974, - 16,919 – 19,152 – 23,193 – 24,181 – 26,281

  Write the claim     mathematically and identify

Which of the following correctly states ?

 Answer:

Find the     critical value(s) and identify the rejection region(s).

What are the critical values? Answer: -1.833, 1.833

Which of the following graphs best depicts the rejection region for this problem?

  Find the     standardized test statistics.     t = -0.60

Decide whether     to reject or fail to reject the null hypothesis.     reject because the test statistics is in the rejection region.

fail to reject     because the test statistic is not in the rejection region.     c. reject because the test statistic is not in the rejection region.     d. fail to reject  because the test statistic is in the rejection     region.

(e) Interpret the decision in the context of the original claim. a. there is sufficient evidence to reject the claim that the mean salary is $22,325. b. there is not sufficient evidence to reject the claim that the mean salary is not $22,325. c. there is sufficient evidence to reject the claim that the mean salary is not $22,325. d. there is not sufficient evidence to reject the claim that the mean salary is $22,325.

Hint:

Test statistic, = -0.60

Critical values are obtained from the Student’s t distribution table with d.f. 9 at the significance level 0.10.

Critical values = -1.833, 1.833

Since the test statistic does not fall in the critical region, we do not reject the null hypothesis.

The times per     week a student uses a lab computer are normally distributed, with a mean     of 6.1 hours and a standard deviation of 1.2 hours. A student is randomly     selected. Find the following probabilities.     (a) the probability that the student uses a lab computer less than 5hrs a     week.     (b) the probability that the student uses a lab computer between 6-8 hrs a     week.

(c) the probability that the student uses a lab computer for more than 9 hrs a week.

  = 0.180

= 0.477

= 0.008

Hint:

P (X <5) == P (Z < -0.9167) = 0.180

P (6<X <8) == P (-0.0833 <Z <1.5833) = 0.477

P (X >9) == P (Z >2.4167) = 0.008

Write the null and     alternative hypotheses. Identify which is the claim.     A study claims that the mean survival time for certain cancer patients     treated immediately with chemo and radiation is 13 months.

  Find the     indicated probability using the standard normal distribution.     P(z>-1.58) = 0.9429

Hint:

P (Z > -1.58) = 1 – P (Z < -1.58) = 1 – 0.0571 = 0.9429

The Gallup     Organization contacts 1323 men who are 40-60 years of age and live in the     US and asks whether or not they have seen their family doctor.

What is the population in the study? Answer: Adult men who are 40-60 years old and live in the US.

What is the sample in the study? Answer: The 1323 adult men who are 40-60 years old and live in the US.

  The ages of 10     brides at their first marriage are given below.     4 32.2            33.6            41.2            43.4            37.1        22.7            29.9            30.6        30.8

(a) find the range of the data set. Range = 20.7 (b) change 43.4 to 58.6 and find the range of the new date set. Range = 35.9 (c) compare your answer to part (a) with your answer to part (b) Answer: Changing the maximum value of the date set greatly affects the range.

 Hint:

  Range = maximum     value – minimum value = 43.4 – 22.7 = 20.7

Range = maximum     value – minimum value = 58.6 – 22.7 = 35.9

  The following appear     on a physician’s intake form. Identify the level of measurement of the     data.     (a) Martial Status     (b) Pain Level (0-10)     (c) Year of Birth     (d) Height

(a) what is the level of measurement for marital status Nominal

 (b)what is the level of measurement for pain level Ordinal

 (c) what is the level of measurement for year of birth Interval

 (d) What is the level of measurement for height Ratio

  To determine her     air quality, Miranda divides up her day into 3 parts; morning, afternoon,     and evening. She then measures her air quality at 3 randomly selected     times during each part of the day. What type of sampling is used?     Answer: Stratified

  Find the     equation of the regression line for the given data. Then construct a     scatter plot of the data and draw the regression line. Then use the     regression equation to predict the value of y for each of the given     x-values, if meaningful. The caloric content and the sodium content (in     milligrams) for 6 beef hot dogs are shown in the table below.

  X= 150 calories

X= 100 calories

X = 120 calories

X = 60 calories

Find the regression equation. = 2.509x + (37.832) Choose the correct graph below.

(a) predict the value of y for x = 150. Answer: 414.182 (b) predict the value of y for x = 100. Answer: 188.372 (c) predict the value of y for x = 120. Answer: 338.912 (d) predict the value of y for x = 60. Answer: not meaningful.

Hint:

Regression equation is y = 2.509x + 37.832 (Please see the excel spreadsheet)

(a) When x = 150, y = (2.509 * 150) + 37.832 = 414.182

(b) When x = 100, y = (2.509 * 100) + 37.832 = 288.732

(c) When x = 120, y = (2.509 * 120) + 37.832 = 338.912

(d) since the value of x is outside the range of values of x, not meaningful

A restaurant     association says the typical household spends a mean of $4072 per year on     food away from home. You are a consumer reporter for a national     publication and want to test this claim. You randomly select 12 households     and find out how much each spent on food away from home per year. Can you     reject the restaurant association’s claim at a = 0.10? Complete parts a     through d.

Write the claim     mathematically and identify. Choose the correct the answer below.

  Use technology to find the P-value. P = 0.03 Decide whether to reject or fail the null hypothesis.

Answer: Reject Interpret the decision in the context of the original claim. Assume the population is normally distributed. Choose the correct answer below. Answer: At the 10% significance level, there is a sufficient evidence to reject the claim.

 Hint: Please see the excel spreadsheet

  The table below     shows the results of a survey in which 147 families were asked if they own     a computer and if they will be taking a summer vacation this year.

  (a) find the probability that a randomly selected family is not taking a summer vacation year. Probability = 0.299 (b) find the probability that a randomly selected family owns a computer Probability = 0.388 (c) find the probability that a randomly selected family is taking a summer vacation this year and owns a computer Probability = 0.825 (d) find the probability a randomly selected family is taking a summer vacation this year and owns a computer. Probability = 0.320

Are the events     of owning a computer and taking a summer vacation this year independent or     dependent events?     Answer: Dependent

Hint:

(a) P (a randomly selected family is not taking a summer vacation year)

=

= 44/147

= 0.299

(b) P (randomly selected family owns a computer) =

= 57/147

= 0.388

(c) P (a randomly selected family is taking a summer vacation this year GIVEN THAT owns a computer)

= (47/147)/(57/147)

= 47/57

= 0.825

(d) P (a randomly selected family is taking a summer vacation this year and owns a computer)

= 47/147

= 0.320

Assume the     Poisson distribution applies. Use the given mean to find the indicated     probability.     Find P(5) when ᶙ = 4

P(5) = 0.156

Hint:

P (5) = , Where λ = 4, x = 5

Therefore, P (5) = = 0.156

In a survey of     7000 women, 4431 say they change their nail polish once a week. Construct     a 99% confidence interval for the population proportion of women who     change their nail polish once a week.

A 99% confidence interval for the population proportion is 618, 0.648

 Hint:

90% Confidence Interval for proportion is given by

 where p = x/n = 4431/7000 = 0.633, = 2.5758, n = 7000

That is,

= (0.618, 0.648)

A random sample     of 53 200-meter swims has a mean time of 3.32 minutes and the population     standard deviation is 0.06 minutes. Construct a 90% confidence interval     for the population mean time. Interpret the results.

The 90% confidence interval is 3.31, 3.33.

Interpret these results. Choose the correct answer: Answer: With 90% confidence, it can be said that the population mean time is between the end points of the given confidence interval.

Hint:

The confidence interval is given by, where = 3.32, σ = 0.06, n = 53, = 1.6449

That is,

= (3.31, 3.33)

  Determine     whether the variable is qualitative or quantitative: Weight

Quantitative Qualitative

32% of college     students say that they use credit cards because of the reward program. You     randomly select 10 college students and ask each to name the reason he or     she uses credit cards. Find the probability that the number of college     students who say they use credit cards because of the reward program is     (a) exactly two, (b), more than two, and (c), between two and five     inclusive.

(a) P(2) = 0.211 (b) P(X>2) = 0.669 (c) P(2<x<5) = 0.816

Hint:

Let X be the number of college students say that they use credit cards because of the reward program. Clearly X is binomial with n = 10 and p = 0.32.  The probability mass function binomial variable is given by .The probability for different values of x are given below.

X

P(X)

P(<=X)

P(<X)

P(>X)

P(>=X)

00.021139228

0.021139228

00.978860772

1

1

0.099478721

0.120617949

0.021139228

0.879382051

0.978860772

2

0.210660821

0.33127877

0.120617949

0.66872123

0.879382051

3

0.264358677

0.595637447

0.33127877

0.404362553

0.66872123

4

0.217707146

0.813344593

0.595637447

0.186655407

0.404362553

5

0.122940506

0.936285099

0.813344593

0.063714901

0.186655407

6

0.048211963

0.984497062

0.936285099

0.015502938

0.063714901

7

0.012964562

0.997461623

0.984497062

0.002538377

0.015502938

8

0.002287864

0.999749487

0.997461623

0.000250513

0.002538377

9

0.000239254

0.999988741

0.999749487

1.1259E-05

0.000250513

10

1.1259E-05

1

0.999988741

01.1259E-05

 (a) P (2) = = 0.211

(b) P (X > 2) = 1 – P (X ≤ 2) = 1 – [P (X = 0) + P (X = 1) + P (X = 2)] = 0.669

(c) P (2 ≤ X ≤ 5) = P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5)

= 0.816

A light bulb     manufacturer guarantees that the mean life of a certain type of light bulb     is at least 950 hours. A random sample of 74 light bulbs has a mean life     of 943 hours with a standard deviation of 90 hours. Do you have enough     evidence to reject the manufacturer’s claim? Use ᶏ     = 0.04

Identify the critical value(s). -1.75 (c) identify the standardized test statistic. z = -0.67 (d) decide whether to reject or fail to reject the null hypothesis.

   6!����R�