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Math 221 Homework Week 7 - LATEST

Math 221 Homework Week 7 - 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=3168&action=edit IF You Face Any Problem Then E Mail Us At [email protected] 1. Math 221 Homework Week 7 - LATEST 2. 3. 4. 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

2. 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

3. 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.

4. 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.

5. 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 > 6. 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.

7. 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 8. Find the critical z values. Assume that the normal distribution applies. Right-tailed test, a = 08. Z = 1.41 9. 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: 10. 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. 11. 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. 12. 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.

13. 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.

14. 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.

15. 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. 16. 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. 17. 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. 18. 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.

19. (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. 20. 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.

#Math 221 Homework Week 7 - LATEST

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MKT 421 Week 1 Complete - New

MKT 421 Week 1 Complete - New 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=934&action=edit IF You Face Any Problem Then E Mail Us At [email protected] MKT 421 Week 1 Complete - New

MKT 421 Week 1 Individual Favorite Brand Paper MKT 421 Week 1 DQ 1 MKT 421 Week 1 DQ 2 MKT 421 Week 1 DQ 3

#MKT 421 Week 1 Complete - New

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MATH 221 Homework Week 1 -Latest

MATH 221 Homework Week 1 -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=3308&action=edit IF You Face Any Problem Then E Mail Us At [email protected] MATH 221 Homework Week 1

1. Determine whether the data set is a population or a sample. Explain your reasoning. The age of each resident in an apartment building. Choose the correct answer below. 1. Population, because it is a collection of ages for all people in the apartment building. 2. Sample, because it is a collection of ages for all people in the apartment building, but there are other apartment buildings 3. Population, because it is a subset of all apartment buildings in the city. 4. Sample, because it is a collection of ages for some people in the apartment building.

1. Determine whether the data set is a population or a sample. Explain your reasoning. The salary of each baseball player in a league Choose the correct answer below 1. Sample, because it is a collection of salaries for some baseball players in the league 2. Sample, because it is a collection of salaries for all baseball players in the league, but there are other sports 3. Population, because it is a collection of salaries for all baseball players in the league 4. Population, because it is a subset of all athletes

2. Determine whether the underlined value is a parameter or a statistic. The average age of men who have walked on the moon was 39 years, 11 months, 15 days. Is the value a parameter or a statistic? Parameter Statistic

2 Determine whether the data set is a population or a sample. Explain your reasoning. The number of pets for 20 households in a town of 300 households. Choose the correct answer below. 1. Sample, because the collection of the number of pets for 20 households is a subset of all households in the town. 2. Population, because it is a collection of the number of pets for all households in the town. 3. Sample, because it is a collection of the number of pets for all households in the town, but there are other towns. 4. Population, because it is a subset of all households in the town.

3. Determine whether the given value is a statistic or a parameter. In a study of all 3336 professors at a college, it is found that 55% own a vehicle. Choose the correct statement below. 1. Statistic because the value is a numerical measurement describing a characteristic of a sample. 2. Statistic because the value is a numerical measurement describing a characteristic of a population. 3. Parameter because the value is a numerical measurement describing a characteristic of a population. 4. Parameter because the value is a numerical measurement describing a characteristic of a sample.

3 Determine whether the underlined value is a parameter or a statistic. In a national survey of high school students (grades 9-12), 25% or respondents reported that someone had offered, sold, or given them an illegal drug on school property. 1. Parameter 2. Statistic 3. Determine whether the given value is a statistic or a parameter. In a study of all 4901 professors at a college, it is found that 35% own a television. Choose the correct statement below. 1. Parameter because the value is a numerical measurement describing a characteristic of a sample. 2. Parameter because the value is a numerical measurement describing a characteristic of a population. 3. Statistic because the value is a numerical measurement describing a characteristic of a sample. 4. Statistic because the value is a numerical measurement describing a characteristic of a population.

4 Determine whether the given value is a parameter or a statistic. In a study of all 1290 employees at a college, it is found that 40% own a computer. Choose the correct statement below. 1. Parameter because the value is a numerical measurement describing a characteristic of a sample. 2. Statistic because the value is a numerical measurement describing a characteristic of a sample. 3. Statistic because the value is a numerical measurement describing a characteristic of a population. 4. Parameter because the value is a numerical measurement describing a characteristic of a population.

5. Determine whether the variable is qualitative or quantitative. Favorite film Is the variable qualitative or quantitative? 1. Qualitative 2. Quantitative

5 Determine whether the given value is a statistic or a parameter. A sample of employees is selected and it is found that 45% own a vehicle. Choose the correct statement below. 1. Parameter because the value is a numerical measurement describing a characteristic of a sample. 2. Statistic because the value is a numerical measurement describing a characteristic of a population. 3. Statistic because the value is a numerical measurement describing a characteristic of a sample. 4. Parameter because the value is a numerical measurement describing a characteristic of a population.

6. Determine whether the variable is qualitative or quantitative. Hair color Is the variable qualitative or quantitative? 1. Qualitative 2. Quantitative

• Determine whether the variable is qualitative or quantitative. Favorite sport Is the variable qualitative or quantitative? 1. Qualitative 2. Quantitative

7. The regions of a country with the six highest per capita incomes last year are shown below. 8. Southeast Western 3 Eastern 4 Northeast 5 Southeast 6 Northern Determine whether the data are qualitative or quantitative and identify the data set’s level of measurement. 1. Qualitative 2. Quantitative What is the data set’s level of measurement? 1. Ratio 2. Ordinal 3. Interval 4. Nominal

• Determine whether the variable is qualitative or quantitative. Car license Is the variable Quantitative? A. Quantitative B Qualitative

8. Which method of data collection should be used to collect data for the following study. The average age of the 105 residents of a retirement community. Choose the correct answer below. 1. Stratified Sampling 2. Census 3. Systematic Sampling 4. Cluster Sampling

• The region representing the top salesperson is a corporation for the past six years is shown below. Northern Northern Eastern Southeast Eastern Northern

Determine whether the data are qualitative or quantitative and identify the data set’s level of measurement.

Are the data qualitative or quantitative? 1. Qualitative 2. Quantitative

What is the data set’s level of measurement? 1. Ratio 2. Nominal 3. Interval 4. Ordinal

9. Decide which method of data collection you would use to collect data for the study. A study of the effect on the taste of a popular soda made with a caffeine substitute. Choose the correct answer below. 1. Observational Study 2. Survey 3. Experiment 4. Simulation

9 Which method of data collection should be used to collect data for the following study. The average weight of 188 students in a high school. Choose the correct answer below. 1. Census 2. Cluster Sampling 3. Stratified Sampling 4. Systematic Sampling

10. Microsoft wants to administer a satisfaction survey to its customers. Using their customer database, the company randomly selects 60 customers and asks them about their level of satisfaction with the company. What type of sampling is used? 1. Simple random 2. Stratified 3. Systematic 4. Cluster 5. Convenience

10 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 snack food made with a sugar substitute. Choose the correct answer below. 1. Survey 2. Simulation 3. Experiment 4. Observational Study

11. A newspaper asks its readers to call in their opinion regarding the number of books they have read this month. What type sampling is used? 1. Simple random 2. Stratified 3. Systematic 4. Cluster 5. Convenience

11 General Motors wants to administer a satisfaction survey to its current customers. Using their customer database, the company randomly selects 80 customers and asks them about their level of satisfaction with the company. What type of sampling is used? 1. Cluster 2. Stratified 3. Convenience 4. Simple random 5. Systematic

12. Determine whether you would take a census or a sampling to collect data for the study described below. The most popular chain restaurant among the 60,000 employees of a company. Would you take a census or use a sampling? 1. Sampling 2. Census

12 A magazine asks its readers to call in their opinion regarding the quality of the articles. What type of sampling is used? 1. Simple random 2. Cluster 3. Systematic 4. Stratified 5. Convenience

13. Math the plot with a possible description of the sample. Choose the correct answer below. 1. Top speeds (in miles per hour) of a sample of sports cars 2. Time (in minutes) it takes a sample of employees to drive to work 3. Grade point averages of a sample of students with finance majors 4. Ages (in years) of a sample of residents of a retirement home

13 Determine whether you would take a census or use a sampling to collect data for the study described below. The most popular house color among the 40,000 employees of a company. Would you take a census or use a sampling? 1. Sampling 2. Census

14. Use a stem-and-leaf plot to display the data. The data represent the heights of eruptions by geyser. What can you conclude about the data? 108 90 110 150 140 120 100 130 110 100 118 106 98 102 105 120 111 130 96 124

Choose the correct stem-and-leaf plot. (Key: 15 ǀ 5 = 155)

A B C

What can you conclude about the data? 12. It appears that most eruptions have a height of around 12. 13. It appears that most eruptions have a height of around 120 14. It appears that most eruptions have a height less than 135 15. It appears that most eruptions have a height greater than 110 16. It appears that most eruptions have a height between 11 and 13

14 Match the plot with a possible description of the sample. Choose the correct answer below. 1. Fastest serve (in miles per hour) of a sample of top tennis players 2. Grade point averages of a sample of students with finance major 3. Time (in minutes) it takes a sample of employees to drive to work 4. Ages (in years) of a sample of residents of a retirement home

15. Determine whether the approximate shape of the distribution in the histogram is symmetric, uniform, skewed left, skewed right, or none of these. Choose the best answer below. 1. Skewed right 2. Skewed left 3. Symmetric 4. Uniform 5. None of these

15 Use a stem-and-leaf plot to display the data. The data represent the heights of eruptions by a geyser. What can you conclude about the data? 106 90 110 150 140 120 100 130 110 101 115 100 99 107 103 120 115 130 95 121

What can you conclude about the data? 1. It appears that most eruptions have a height of around 12 2. It appears that most eruptions have a height less than 135 3. It appears that most eruptions have a height greater than 110 4. It appears that most eruptions have a height between 11 and 13 5. It appears that most eruptions have a height of around 120

16. The maximum number of seats in a sample of 13 sport utility vehicles are listed below. Find the mean, median, and mode of the data. 5 7 8 8 5 6 4 4 4 4 4 4 5 Find the mean. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 5. The mean is 2 (Type the integer or decimal rounded to the nearest tenth as needed) 1. The data does not have a mean.

Find the median. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 1. The median is 5 (Type the integer or decimal rounded to the nearest tenth as needed) 1. The data does not have a median.

Find the mode. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 1. The median is 4 (Type the integer or decimal rounded to the nearest tenth as needed) 1. The data does not have a mode.

16 Determine whether the approximate shape of the distribution in the histogram is symmetric, uniform, skewed left, skewed right, or none of those. Choose the best answer below. 1. Skewed right 2. Skewed left 3. Symmetric 4. Uniform 5. None of these

17. Find the range, mean, variance, and standard deviation of the sample data set. 14 12 13 8 20 7 18 16 15 The range is 13

= 13.7 (round to the nearest tenth as needed) • = 25 S = 4.3

17 The maximum number of seats in a sample of 13 sport utility vehicles are listed below. Find the mean, median, and mode of the data. 8 10 11 11 8 7 7 7 9 7 7 7 8 Find the mean. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 8. The mean is 2 (Type the integer or decimal rounded to the nearest tenth as needed) 1. The data does not have a mean. Find the median. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 1. The median is 8 (Type the integer or decimal rounded to the nearest tenth as needed) 1. The data does not have a median Find the mode. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 1. The mode is 7 (Type the integer or decimal rounded to the nearest tenth as needed) 1. The data does not have a mode

18. The ages of 10 brides at their first marriage are given below. 35.9 32.6 28.7 37.7 44.7 31.3 29.5 23.2 22.4 33.6 (a) Find the range of the data set. Range = 22.3 (round to the nearest tenth as needed) (b) Change 44.7 to 61.3 and find the range of the new data set. Range = 38.9 (round to the nearest tenth as needed) (c) Compare your answer to part (a) with your answer to part (b)

1. Changing the maximum value of the data set does not affect the range 2. Changing the minimum value of the data set does not affect the range 3. Changing the minimum value of the data set greatly affects the range 4. Changing the maximum value of the data set greatly affects the range

18 Find the range, mean, variance, and standard deviation of the sample data set. 10 14 13 5 19 11 18 12 9 The range is14

19. Heights of men on a baseball team have a bell-shaped distribution with a mean of 185 cm and a standard deviation of 6 cm. Using the empirical rule, what is the approximate percentage of the men between the following values? 20. 173 cm and 197 cm 21. 167 cm and 203 cm

A 95% of the men are between 173 cm and 197 cm B 99.7% of the men are between 167 cm and 203 cm (Do not round)

20. The mean value of land and buildings per acre from a sample of farms is $1400, with a standard deviation of $100. The data set has a bell-shaped distribution. Assume the number of farms in the sample is 70. • Use the empirical rule to estimate the number of farms whose land and building values per acre are between $1300 and $1500. 48 farms (Round to the nearest whole number as needed) • If 28 additional farms were sampled, about how many of these additional farms would you expect to have land and building value between $1300 per acre and $1500 per acre? 19 farms (Round to the nearest whole number as needed)

21. Use the box-and-whisker plot to identify • The minimum entry • The maximum entry • The first quartile • The second quartile • The third quartile • The interquartile range

• Min = 11 • Max = 21 • Q 1 = 14 • Q 2 = 16 • Q 3 = 18 • IQR = 4

• The ages of 10 brides at their first marriage are given below 22.6 23.735.6 39.2 44.7 29.6 30.8 31.7 24.5 28.2 (a) Find the range of the data set Range =22.1 (Round to the nearest tenth as needed) B Change 44.7 to 68.3 and find the range of the new data set Range = 45.7 (Round to the nearest tenth as needed)

1. Compare your answer to part (a) with your answer to part (b). • Changing the maximum value of the data set greatly affects the range • Changing the minimum value of the data set greatly affects the range • Changing the maximum value of the data set does not affect the range • Changing the minimum value of the data set does not affect the range

20 Heights of men on a baseball team have a bell-shaped distribution with a mean of 172 cm and a standard deviation of 7 cm. Using the empirical rule, what is the approximate percentage of the men between the following values? 1. 151 cm and 193 cm 2. 165 cm and 179 cm

99. 7% of the men are between 151 cm and 193 cm (Do not round) 100. 68 % of the men are between 165 cm and 179 cm (Do not round)

20. The midpoints A, B, and C are marked on the histogram. Match them to the indicated scores. Which scores, if any, would be considered unusual?

The point A corresponds with z = -2.17 The point B corresponds with z = 0 The point C corresponds with z = 1.41 Which scores, if any, would be considered unusual? 1. 41 2. -2.17 3. 0 4. None

#MATH 221 Homework Week 1 -Latest

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Fin 370

Fin 370 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=3176&action=edit IF You Face Any Problem Then E Mail Us At [email protected] 1. 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? 1. 2 Find the indicated z-score shown in the graph to the right. 1. 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.

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 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?

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

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.

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)

Find the margin of error for the given values of c,s, and n. Find the critical value Tc for the confidence level c = .90 and sample size n = 29. 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

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).

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

Find the indicated probability using the standard normal distribution.

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

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.

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

1. 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. 1. What is the total area under the normal curve? \ 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

#Fin 370

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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 [email protected] 1. 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

2. 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

3. 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. 4. 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

4. 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 5. 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 6. 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.

7. 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

8. 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.

9. 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

10. 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 11. 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 12. 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

13. 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

14. 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 15. 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

16. 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 17. 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.

18. 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%. 19. 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

20. What is the total area under the normal curve? Answer = 1 21. 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 22. 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. 23. 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

#DeVry MATH 221 Week 7 Quiz with Formulas – Latest

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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 [email protected] 1. DeVry MATH 221 Week 7 Homework - Latest 2015 2. 3. 4. 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

#DeVry MATH 221 Week 7 Homework - Latest

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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 [email protected] 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?

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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 [email protected] 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.

1. 90% 2. 98% 3. 99% 4. 95%

1. Given the same sample statistics, which level of confidence would produce the widest confidence interval? Choose the correct answer below. 1. 99% 2. 95% 3. 98% 4. 90%

Use the values on the number line to find the sampling error. The sampling error is 1.56

2. 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. 1. 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.

1. The 90% confidence interval 2. 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? 1. The 95% confidence interval 2. 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?

1. The sample size is the same for all levels of confidence 2. The lower the confidence level the larger the sample size 3. 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?

1. The t-distribution has the wider interval 2. 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

1. The interval from part (a), which uses the normal distribution, is narrower than the interval from part (b), which uses the t-distribution. 2. The interval from part (a), which uses the t-distribution, is wider than the interval from part (b), which uses the normal distribution. 3. The interval from part (a), which uses the normal distribution, is wider than the interval from part (b), which uses the t-distribution. 4. 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.

1. 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. 2. 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. 3. 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?

1. The 90% confidence interval 2. 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.)

1. 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.)

#DeVry MATH 221 Week 6 Homework - Latest

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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 [email protected] 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.

#DeVry MATH 221 Week 6 DQ Confidence Interval Concepts - Latest

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DeVry MATH 221 Week 5 Homework - Latest

DeVry MATH 221 Week 5 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=3324&action=edit IF You Face Any Problem Then E Mail Us At [email protected] MATH 221 Homework Week 5

1 A study was conducted that resulted in the following relative frequency histogram. Determine whether or not the histogram indicates a normal distribution could be used a model for the variable.

1. The histogram is not bell-shaped, so a normal distribution could not be used as a model for the variable. 2. The histogram is bell-shaped, so a normal distribution could be used as a model for the variable. 3. The histogram is not bell-shaped, so a normal distribution could be used as a model for the variable. 4. The histogram is bell-shaped, so a normal distribution could not be used as a model for the variable.

1. A study was conducted that resulted in the following relative frequency histogram. Determine whether or not the histogram indicates a normal distribution could be used a model for the variable. 1. The histogram is not bell-shaped, so a normal distribution could not be used as a model for the variable. 2. The histogram is bell-shaped, so a normal distribution could be used as a model for the variable. 3. The histogram is not bell-shaped, so a normal distribution could be used as a model for the variable. 4. The histogram is bell-shaped, so a normal distribution could not be used as a model for the variable.

2 Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1.

The area of the shaded region is 0.7823 (round to 4 decimal places as needed.)

2. Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. The area of the shaded region is .7291 (round to 4 decimal places as needed.)

3. Find the area of the indicated region under the standard normal curve. The area between z = 0 and z = 1 under the standard normal curve is .3413. (round to 4 decimal places as needed.)

3 Find the area of the indicated region under the standard normal curve.

The area between z = 0 and z = 1.3 under the standard normal curve is 0.4032. (round to 4 decimal places as needed.)

4. Find the indicated area under the standard normal curve. To the left of z = -0.28 The area to the left of z = -0.28 under the standard normal curve is .3897. (round to 4 decimal places as needed.) 5. Find the indicated area under the standard normal curve. To the left of z = - 0.28 The area to the left of z = - 0.28 under the standard normal curve is .3897. round to four decimal places as needed. 6. Find the indicated area under the normal curve. between z = -1.08 and z = 1.08 The area between z= -1.08 and z = 1.08 under the standard normal curve is .7198 (round to 4 decimal places as needed.)

6 Find the indicated area under the standard normal curve. To the left of z = 2.71

The area to the left of z = 2.71 under the standard curve is .9966.

6 Find the indicated area under the standard normal curve. To the right of z = -0.31

The area to the right of z = -0.31 under the standard normal curve is .6217.

6 Find the indicated area under the standard normal curve. Between z = -2.82 and z = 2.82 The area between z = -2.82 and z = 2.82 under the standard normal curve is .9952. 7 Assume the random variable x is normally distributed with mean µ = 82 and standard deviation . Find the indicated probability. P(x<75) P(x<75) = .0401 (Round to four decimal places as needed)

7. Assume random variable x is normally distributed with mean µ = 82 and standard deviation. Find the indicated probability. P(x<78) P(X<78) = .1587 (round to 4 decimal places as needed. 8. Assume random variable x is normally distributed with mean µ = 87 and standard deviation . Find the indicated probability. P(75<x<84) P(75<x<84) = .2253 (round to 4 decimal places as needed.)

8 Assume the random variable x is normally distributed with mean µ = 84 and standard deviation σ = 5. Find the indicated probability. P(68<x<74) P(68<x74<) = 0.0221 (Round to four decimal places as needed)

9 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 67.5 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts (a) through (c).

1. Find the probability that his height is less than 67 inches. The probability that the study participant selected at random is less than 67 inches tall is 0.4013. (Round to four decimal places as needed) 1. Find the probability that his height is between 67 and 72 inches. The probability that the study participant selected at random is between 67 inches and 72 inches tall is 0.5865. (Round to four decimal places as needed) 1. Find the probability that his height is more than 72 inches. The probability that the study participant selected at random is more than 72 inches tall is 0.0122. (Round to four decimal places as needed)

9. A survey was conducted to measure the height of men. In the survey, respondents were grouped by age. In the 20-29 group, the heights were normally distributed, with a mean of 69.9 inches and a standard deviation of 3.0 inches. A study participant is randomly selected. Complete parts (a) through (c). find the probability that his height is less than 68 inches. The probability that the student participant selected at random is less than 68 inches tall is .2644 (round to 4 decimal places as needed.) B. find the probability that his height is between 68 and 71 inches the probability that the student participant selected at random is between 68 and 71 inches tall is ? (round to 4 decimal places as needed.) 10. find the probability that his height is more than 71 inches. The probability that the study participant selected at random is more than 71 inches tall is .3557. (round to 4 decimal places as needed).

10 Use the normal distribution of SAT critical reading scores for which the mean is 515 nad the standard deviation is 108. Assume the variable x is normally distributed. • What percent of the SAT verbal scores are less than 650? • If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 575?

89. Approximately 44% of the SAT verbal scores are less than 650. (Round to two deciam places as need) 90. You would expect that approximately 289 SAT verbal scores would be greater than 575. (Round to the nearest whole number as needed)

10. Use the normal distribution of Sat critical reading scores for which the mean is 514 and the standard deviation is 124. Assume the variable x is normally distributed. what percent of the SAT verbal score is less than 500? B. if 1000 sat scores are randomly selected, how many would be greater than 525? A: 61.41% B: 464 greater than 525

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

The z-score is -1.13 (Round to two decimal places as needed)

11 Find the indicated z-score shown in the graph to the right. The z-score is .78.

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

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

The z-score is -.22.

The z-score is 1.20 (Round to two decimal places as needed)

13 Find the z-score that has 11.9% of the distribution’s area to its right. The z-score is 1.18 (Round to two decimal places as needed) 13. Find the z-score that has 11.9% of the distribution’s area to its right.

The z-score is 1.18 (Round to two decimal places as needed)

14 Find the z-scores for which 70% of the distribution’s area lies between –z and z. The z-scores are -1.04,1.04 (Use a comma to separate answers as needed. Round to two decimal places as needed)

14. Find the z-scores for which 98% of the distribution’s area lies between – z and z.

The z-scores are – 2.32,2.32 (Use a comma to separate answers as needed)

15 In a survey of women in a certain country (ages 20-29), the mean height was 66.5 inches with a standard deviation of 2.84 inches. Answer the following questions about the specified normal distribution. • What height represents the 85th percentile? • What height represents the first quartile? 69. The height the represents the 85th percentile is 44 inches. (Round to two decimal places as needed) 70. The height that represents the first quartile is 58 inches. (Round to two decimal places as needed) 71.

15. In a survey of women in a certain country (ages 20-29), the mean height was 65.9 inches with a standard deviation of 2.74 inches. Answer the following questions about the specified normal distribution. • What height represents the 95th percentile? • What height represents the first quartile?

• The height that represents the 95th percentile is 42 inches. (Round to two decimal places as needed) • The height that represents the first quartile is 06 inches. (Round to two decimal places as needed)

16 The time spent (in days) waiting for a heart transplant in two states for patients with type A + blood can be approximately by a normal distribution, as shown in the graph. Complete parts (a) and (b) below.

• What is the shortest time spent waiting for a heart that would still place a patient in the top 30% of waiting times? 146.64 days (Round to two decimal places as needed) (b) What is the longest time spent waiting for a heart that would still place a patient in the bottom 29% of waiting times? 116.32 days. (Round to two decimal places as needed)

16. The time spent (in days) waiting for a heart transplant in two states for patients with type A + blood can be approximated by a normal distribution, as shown in the graph to the right. Complete parts (a) and (b) below.

• What is the shortest time spent waiting for a heart that would still place a patient in the top 15% or waiting times? 152.34 days ( Round to two decimal places as needed) (b) What is the longest time spent waiting for a heart that would still place a patient in the bottom 5% of waiting times? 102.4 days ( Round to two decimal places as needed)

17 population has a mean µ = 86 and a standard deviation σ = 14. Find the mean and standard deviation of a sampling distibution of a sampling distribution of sample means with sample size n = 49.

= 86(Simplify your answer) = 2(Simplify your answer)

17. A population has a mean µ = 81 and a standard deviation Find the mean and standard deviation of a sampling distribution of sample means with sample size n = 49.

= 81(Simplify your answer) = 1(Simplify your answer)

18. The graph of the waiting time (in seconds) at a red light is shown below on the left with its mean and standard deviation. Assume that a sample size of 100 is drawn from the population. Decide which of the graphs labeled (a)-(c) would most closely resemble the sampling distribution of the sample means. Explain your reasoning.

Graph (c) most closely resembles the sampling distribution of the sample means, because 17.7, 1.16, and the graph approximates a normal curve. (Type an integer or a decimal)

18 The graph of the waiting time (in seconds) at a red light is shown below on the left with its mean and standard deviation. Assume that a sample size of 100 is drawn from the population. Decide which of the graphs labeled (a)-(c) would most closely resemble the sampling distribution of the sample means. Explain your reasoning.

Graph (a) most closely resembles the sampling distribution of the sample means, because µ - = 17.4, σ - = 1.27, and the graph approximates a normal curve. X x (Type an integer or a decimal)

19 A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 133 ounces and a standard deviation of 0.30 ounce. You randomly select 50 cans and carefully measure the contents. The sample mean of the cans is 132.9 ounces. Does the machine need to be reset? Explain your reasoning.

Yes, it is very unlikely that you would have randomly sampled 50 cans with a mean equal to 132.9 ounces, because it does not lie within the range of a usual event, namely within 2 standard deviation of the sample means.

19. A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 132 ounces and a standard deviation of 0.30 ounces. You randomly select 45 cans and carefully measure the contents. The sample mean of the cans is 131.9 ounces. Does the machine need to be reset? Explain your reasoning. Yes, it is very unlikely that you would have randomly sampled 45 cans with a mean equal to 131.9 ounces, because it does not lie within the range of a usual event, namely within 2 standarddeviations of the mean of the sample means.

20 A manufacturer claim that the life span of its tires is 50,000 miles. You work for a consumer protection agency and you are testing these tires. Assume the life spans of the tires are normally distributed. You select 100 tires at random and test them. The mean life span is 49,741 miles. Assume σ = 900. Complete parts (a) through (c)

1. Assuming the manufacturer’s claim is correct, what is the probability that the mean of the sample is 49,741 miles of less? 0.0020 (Round to four decimal places as needed) 1. using your answer form part (a), what do you think of the manufacturer’s claim?

The claim is inaccurate because the sample mean would be considered unusual since it does not lie within the range of a usual event, namely within 2 standard deviations of the mean of the sample means.

1. Assuming the manufacturer’s claim is true, would it be unusual to have an individual tire with a life span of 49, 741 mile? Why or why not? No, because 49,741 lies within the range of a usual event, namely within 2 standard deviations of the mean for an individual tire.

20. A manufacturer claims that the life span of its tires is 49,000 miles. You work for a consumer protection agency and you are testing these tires. Assume the life spans of the tires are normally distributed. You select 100 tires at random and test them. The mean life span is 48, 778 miles. Assume σ = 900. Complete parts (a) through (c). • Assuming the manufacturer’s claim is correct, what is the probability that the mean of the sample is 48,778 miles or less?

.0069 (Round to four decimal places as needed)

• Using your answer from part (a), what do you think of the manufacturer’s claim? The claim is inaccurate because the sample mean would be considered unusual since it does not lie within the range of a usual event, namely within 2 standard deviations of the mean of the samples means. ( c) Assuming the manufacturer’s claim is true, would it be unusual to have an individual tire with a life span of 48,778 mile? Why or why not? No, because 48,778 lies within the range of a usual event, namely within 2 standard deviations of the mean for an individual tire.

#DeVry MATH 221 Week 5 Homework - Latest

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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 [email protected] 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?

#DeVry MATH 221 Week 5 DQs Interpreting Normal Distributions - Latest

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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 [email protected] MATH 221 Homework Week 4 - 2015

1. 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 1. Choose the correct answer below. Use histogram 2. X = 0, x = 1, x = 2, x = 3, and x = 4 3. X = 3 and x = 4 4. X = 0 and x = 1 5. There are no unusual values of x in the histogram 6. X = 7 7. X = 0, x = 1, x = 2, x = 3, and x = 4 8. X = 0 and x = 1 9. X = 0 and x = 1 10. There are no unusual values of x in the histogram

2. 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). 1. X = 4 2. X = 0, x =7, and x = 8 3. X = 2 4. There are no unusual values of x in the histogram • Choose the correct answer below. Use the histogram (b) 1. X =0, x =7, and x = 8 2. X = 4 3. X = 4 4. There are no unusual values of x in the histogram

3. 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? 1. No 2. Yes What is a success in this experiment? 1. Baby recovers 2. Baby doesn’t recover 3. 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. 1. N = 6 2. This is not a binomial experiment 3. Specify the value of p. Select the correct choice below and fill in any answer boxes in your choice. 4. p = 0.7 5. This is not a binomial experiment 6. Specify the value of q. Select the correct choice below and fill in any answer boxes in your choice. 7. q = 0.3 8. This is not a binomial experiment List the possible values of the random variable x. 1. X = 0, 1, 2,…,5 2. X = 0, 1, 2,…6 3. X = 1, 2, 3,…6 4. 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. 1. P(8) = .034 (Round to the nearest thousandth as needed) 2. P(x≥8) = 042(Round to the nearest thousandth as needed) 3. 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. 1. 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) 1. 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) 1. 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. 1. 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) 1. 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) 1. 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. 1. P(2) = .205 (Round to the nearest thousandth as needed.) 2. P(X >2) = 626 (Round to the nearest thousandth as needed.) 3. 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. 1. P(2) = 142(Round to the nearest thousandth as needed.) 2. P(X >2) = 798(Round to the nearest thousandth as needed.) 3. 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. 1. Construct a binomial distribution using n = 6 and p = 0.36 X P(x) • .069 • .232 • .326 • .245 • .103 • .023 • .002

1. Choose the correct histogram for this distribution below.

1. Describe the shape of the histogram 2. Skewed right 3. Skewed left 4. Symmetrical 5. None of these 6. 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 1. Skewed right 2. Skewed left 3. Symmetrical 4. 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.

15. 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.)

16. 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.)

17. 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.)

18. 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.)

19. 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? 1. Poisson distribution 2. Binomial distribution 3. Geometric distribution

20. 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? 21. You are interested in counting the number of successes out of n trials. 22. You are interested in counting the number of occurrences that take place within a given unit of time. 23. You are interested in counting the number of trials until the first success.

21. 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?

1. Binomial distribution 2. Geometric distribution 3. Poisson distribution

22. 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. 1. You are interested in counting the number of occurrences that take place within a given unit of time. 2. You are interested in counting the number of successes out of n trials. 3. You are interested in counting the number of trials until the first success.

23. 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.) 1. b) P(make your sale on the first, second, or third call) = 364 (Round to three decimal places as needed.) 1. 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. 1. The event in part (a), “make your first sale on the fifth call”, is unusual 2. The event in part (b), “make you sale on the first, second, or third call”, is unusual 3. The event in part ©, “do not make a sale on the first three calls”, is unusual 4. None of the events are unusual

24. 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.) 25. P(at most five typographical errors are found on a page) = 7851 (Round to four decimal places as needed.) 1. 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. 1. The event in part (a) is unusual. 2. The event in part (b) is unusual. 3. The event in part (c) is unusual. 4. None of the events are unusual

25. 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.) 1. 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. 1. The event in part (a) is unusual. 2. The event in part (b) is unusual. 3. The event in part (c) is unusual. None of the events are unusual

#DeVry MATH 221 Week 4 Homework - Latest

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