A university with a high water bill is interested in estimating the mean amount of time that students spend in the shower each day. In a sample of 12 students, the average time was 4.75 minutes and the standard deviation was 1.26 minutes. Using this sample information, construct a 99% confidence interval for the mean amount of time that students spend in the shower each day. Assume normality.

Step-by-step explanation:

We want to determine a 99% confidence interval for the mean amount of time that students spend in the shower each day.

Number of sample, n = 12

Mean, u = 4.75 minutes

Standard deviation, s = 1.26 minutes

For a confidence level of 99%, the corresponding z value is 2.58. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

4.75 ± 2.58 × 1.26/√12

= 4.75 ± 2.58 × 0.364

= 4.75 ± 0.939

The lower end of the confidence interval is 4.75 - 0.939 = 3.811

The upper end of the confidence interval is 4.75 + 0.939 =5.689

Therefore, with 99% confidence interval, the mean amount of time that students spend in the shower each day is between 3.811 minutes and 5.689 minutes

inequality:

620-37.4p ≥ 450

the minimum amount of packages needed to be removed from the elevator is 5.

step-by-step explanation: