Question

The population of men at UMBC has a mean height of 69 inches with a standard deviation of 4 inches. The women at UMBC have a mean height of 65 inches with a standard deviation of 3 inches. A sample of 50 men and 40 women is selected. What is the probability that the sample mean of men heights is more than 5 inches greater than the sample mean of women heights

Answer 1

Answer:

The probability that the sample mean of men heights is more than 5 inches greater than the sample mean of women heights is 0.0885.

Step-by-step explanation:

We are given that the population of men at UMBC has a mean height of 69 inches with a standard deviation of 4 inches. The women at UMBC have a mean height of 65 inches with a standard deviation of 3 inches.

A sample of 50 men and 40 women is selected.

The z-score probability distribution for the two-sample normal distribution is given by;

                          Z  =  $\frac{(\bar X_M-\bar X_W)-(\mu_M-\mu_W)}{\sqrt{\frac{\sigma_M^{2} }{n_M}+\frac{\sigma_W^{2} }{n_W} } }$  ~ N(0,1)

where, $\mu_M$ = population mean height of men at UMBC = 69 inches

           $\mu_W$ = population mean height of women at UMBC = 65 inches

           $\sigma_M$ = standard deviation of men at UMBC = 4 inches

           $\sigma_M$ = standard deviation of women at UMBC = 3 inches

           $n_M$ = sample of men = 50

           $n_W$ = sample of women = 40

Now, the probability that the sample mean of men heights is more than 5 inches greater than the sample mean of women heights is given by = P($\bar X_M-\bar X_W$ > 5 inches)

 P($\bar X_M-\bar X_W$ > 5 inches) = P( $\frac{(\bar X_M-\bar X_W)-(\mu_M-\mu_W)}{\sqrt{\frac{\sigma_M^{2} }{n_M}+\frac{\sigma_W^{2} }{n_W} } }$ > $\frac{(5)-(69-65)}{\sqrt{\frac{4^{2} }{50}+\frac{3^{2} }{40} } }$ ) = P(Z > 1.35)

                                         = 1 - P(Z $\leq$ 1.35) = 1 - 0.9115 = 0.0885

The above probability is calculated by looking at the value of x = 1.35 in the z table which has an area of 0.9115.