Question

Question 1

“PhysicalFun” is a physical center which opens 24 hours per day. Each member of “PhysicalFun” has an access card. Members would be charged in a weekly basic (Sunday 12:00a.m. to Saturday 11: 59 p.m.) according to the accumulated number of hours of facilities used within a week. The weekly basic charge is $150 with an additional charge of $30 per hour. According to the company record, the number of hours a member spends in “PhysicalFun” in a week is normally distributed with mean 15 hours and standard deviation 2.8 hours.

(a) What are the average and standard deviation of weekly charge of a member?

(b) Suppose the middle 85% of weekly charge of a member is denoted by $(L1, L2). Find the values of L1 and L2.

(c) The senior management suggests fixing the weekly charge per member at $700. Assume the number of hours a member spends in “PhysicalFun” would not be changed due to the change of the weekly charge calculation method. What proportion of members would pay more money than the original system?

(d) How many hours does a member spend in “PhysicalFun” in a week so he / she would be beneficial by the new system?

Question 2

The Body Mass Index (BMI) is a value calculated based on the weight and the height of an individual. In a small European city, a survey was conducted one year ago to review the BMI of the citizens. In the sample with 200 citizens, the mean BMI was 23.3 kg/m2 and standard deviation was 1.5 kg/m2 . It is reasonable to assume the BMI distribution is a normal distribution.

(a) Find the point estimate of the population mean BMI one year ago

(b) Calculate the sampling error at 90% confidence.

(c) Construct a 90% confidence interval estimate of the population mean BMI one year ago.

This city launched a healthy exercise program to reduce citizen’s BMI after last year’s survey. Suppose the program effectively reduces the BMI of each citizen by 2.5%.

(d) Construct a 98% confidence interval estimate of the population mean BMI after the healthy exercise program. (Hint: find the updated sample mean and sample standard deviation of the BMI of the sample with 200 citizens selected last year)

Answer 1

Que.2

a. Sample mean is the point estimate of population mean. Therefore population mean is 23.3 kg/m2

b.

Sampling error at 90% confidence interval is,

c.

90% confidence interval one year ago is,

d.

98% confidence interval after healthy exercise program.

Each citizen's BMI is reduced by 2.5 %. Thus mean BMI become

23.3 - (23.3*2.5 / 100 ) = 22.72

We know that there is standard deviation is invariant to change of origin, hence standard deviation remains same.

Where is obtained using t table. It does not contain degrees of freedom 199, hence we use value for 200. Since sample size is very large both the values are approximately same.