Question

1) The normal distribution is the most important continuous distribution in statistics. Give at least three reasons why? 2. The director of a hospital wishes to estimate the mean number of people who are admitted to the emergency room during a 24-hour period. The director randomly selects 49 different 24-hour periods and determines the number of admissions for each. For this sample, x = 17.2 and s2 = 25. Estimate the mean number of admissions per 24-hour period with a 90% confidence interval. 3. A retired statistician was interested in determining the average cost of a $200,000.00 term life insurance policy for a 60-year-old male non-smoker. He randomly sampled 65 subjects (60-year-old male non-smokers) and constructed the following 95 percent confidence interval for the mean cost of the term life insurance: ($850.00, $1050.00). State the appropriate interpretation for this confidence interval. Note that all answers begin with "We are 95 percent confidence that…" 4. How much money does the average professional football fan spend on food at a single football game? That question was posed to 60 randomly selected football fans. The sampled results show that the sample mean was $70.00 and prior sampling indicated that the population standard deviation was $17.50. Use this information to create a 95 percent confidence interval for the population mean. 5. The following data represent the scores of a sample of 50 randomly chosen students on a standardized test. 39 48 55 63 66 68 68 69 70 71 71 71 73 74 76 76 76 77 78 79 79 79 79 80 80 82 83 83 83 85 85 86 86 88 88 88 88 89 89 89 90 91 92 92 93 95 96 97 97 99 3 Write a 95% confidence interval for the mean score of all students who took the test. 6. How many tissues should a package of tissues contain? Researchers have determined that a person uses an average of 41 tissues during a cold. Suppose a random sample of 10,000 people yielded the following data on the number of tissues used during a cold: x = 35, s = 18.

Answer 1

1. The normal distribution is the most important continuous distribution in statistics.

The reasons for the above statement are given as follows

• Many real-life problems consisting of continuous data exhibit a bell-shaped curve when compiled and graphed which means that most of the data is lies near the average value and very few of them lie towards the extreme ones.
• The mean and standard deviation of a normal distribution is given by 0 and 1 respectively indicating that there is very less variation in the data from the mean value

2. The 90% confidence interval for the population mean (with unknown population standard deviation) is given by

• 

  where x-bar is sample mean and s is sample standard deviation

  z alpha is the value of the standard normal distribution

  Thus,

  x-bar= 17.2

  s=25

  n=49

  z alpha =1.645

  Upper Limit: 17.2+ 1.645*(25/sqrt(49)) = 23.05

• Lower Limit: 17.2- 1.1.6456 *(25/sqrt(49)) = 11.32

  Thus, we are 90% confident that the average number of admissions per 24-hour period lie between 11.32 and 23.05

3. We are 95% confident that the mean cost of term life insurance lies between $850 and $1050

4. The 95% confidence interval for the population mean (with known population standard deviation) is given by

Thus ,

x-bar= 80

n=60

sd=17.50

z-value (0.05)= 1.96

CI for Population mean is

Upper Limit: 80 + 1.96 *(17.50 /sqrt(60)) = 84.42

Lower Limit: 80 - 1.96 *(17.50 /sqrt(60)) = 75.57

Thus , we are 95% confident that the average money a professional football fan spend on food at a single football game lies between $75.57 and $84.42

5.The 95% confidence interval for the population mean (with unknown population standard deviation) is given by

where x-bar is sample mean and s is sample standard deviation

z alpha is the value of the standard normal distribution

Thus,

x-bar= 79.98

s=12.34

n=50

z alpha =1.96

Upper Limit: 79.98 + 1.96 *(12.34/sqrt(60)) = 83.40

Lower Limit: 79.98 - 1.96 *(12.34/sqrt(60)) = 76.56

Thus , we are 95% confident that the average test scores lie between 76.56 and 83.40