Question

Weights of men: 90% confidence; n = 14, x=161.5 lb, s =13.7 lb

Answer 1

To calculate the probability that 20 randomly selected men will have a sum weight greater than 3,600 pounds, we can use the Central Limit Theorem and the standard normal distribution table.

Step-by-step explanation:

To calculate the probability that 20 randomly selected men will have a sum weight greater than 3,600 pounds, we can use the Central Limit Theorem. As the sample size is large (n = 20), we can assume that the sample mean follows a normal distribution. The mean weight of men is 161.5 lb and the standard deviation is 13.7 lb. Since the sample mean follows a normal distribution, we can calculate the z-score using the formula: z = (x - μ) / (σ / √n). Once we have the z-score, we can use the standard normal distribution table to find the probability.

Answer 2

The 90% for the average weights of men is between 137.24 lb and 185.76 lb.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 14 - 1 = 14

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.9)/(2) = 0.95`. So we have T = 1.7709

The margin of error is:

M = T*s = 1.7709*13.7 = 24.26

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 161.5 - 24.26 = 137.24 lb

The upper end of the interval is the sample mean added to M. So it is 161.5 + 24.26 = 185.76 lb

The 90% for the average weights of men is between 137.24 lb and 185.76 lb.