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

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.

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{(\sigma_M^(2) )/(n_M)+(\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{(\sigma_M^(2) )/(n_M)+(\sigma_W^(2) )/(n_W) } }` >
`\frac{(5)-(69-65)}{\sqrt{(4^(2) )/(50)+(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.