Question

Assuming that men's weights are normally distributed with a mean of 172 lb. and a population standard deviation of 29 lb, find the probability. a.) That a randomly selected man has a weight greater than 180 lb. (4 points) b.) That 36 randomly selected men have a mean weight of less than 167 lb. (4 points)

Answer 1

a) Normal distribution: P(X < A) = P(Z < (A - $\mu$ )/$\sigma$)

$\mu$ = 172 lb

$\sigma$ = 29 lb

P(a randomly selected man has a weight greater than 180 lb), P(X > 180) = 1 - P(X < 180)

= 1 - P(Z < (180 - 172)/29)

= 1 - P(Z < 0.28)

= 1 - 0.6103

= 0.3897

b) Sample size, n = 36

For sampling distribution of mean,

P(
T
< A) = P(Z < (A - $\mu_\bar{x}$ )/$\sigma_\bar{x}$)

$\mu_\bar{x}$ = $\mu$ = 172 lb

$\sigma_\bar{x}$ =

0/vn

= $29/\sqrt{36}$

= 4.8333

P(36 randomly selected men have a mean weight of less than 167 lb) = P( < 167)

T

= P(Z < (167 - 172)/4.8333)

= P(Z < -1.03)

= 0.1515