Respuesta :
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 = [tex]\frac{(\bar X_M-\bar X_W)-(\mu_M-\mu_W)}{\sqrt{\frac{\sigma_M^{2} }{n_M}+\frac{\sigma_W^{2} }{n_W} } }[/tex] ~ N(0,1)
where, [tex]\mu_M[/tex] = population mean height of men at UMBC = 69 inches
[tex]\mu_W[/tex] = population mean height of women at UMBC = 65 inches
[tex]\sigma_M[/tex] = standard deviation of men at UMBC = 4 inches
[tex]\sigma_M[/tex] = standard deviation of women at UMBC = 3 inches
[tex]n_M[/tex] = sample of men = 50
[tex]n_W[/tex] = 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([tex]\bar X_M-\bar X_W[/tex] > 5 inches)
P([tex]\bar X_M-\bar X_W[/tex] > 5 inches) = P( [tex]\frac{(\bar X_M-\bar X_W)-(\mu_M-\mu_W)}{\sqrt{\frac{\sigma_M^{2} }{n_M}+\frac{\sigma_W^{2} }{n_W} } }[/tex] > [tex]\frac{(5)-(69-65)}{\sqrt{\frac{4^{2} }{50}+\frac{3^{2} }{40} } }[/tex] ) = P(Z > 1.35)
= 1 - P(Z [tex]\leq[/tex] 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.
