Answer:
A. 0.0021
Step-by-step explanation:
Given that the heights of men are normally distributed with a mean of 66.9 inches and a standard deviation of 2.1inches.
Sample size = 36
Std dev of sample = [tex]\frac{2.1}{\sqrt{36} } =0.35[/tex]
The sample entries X the heights are normal with mean= 66.9 inches and std deviation = 0.35 inches
Or we have
Z = [tex]\frac{x-66.9}{0.35}[/tex]
Hence the probability that they have a mean height greater than 67.9 inches
=[tex]P(X>67.9)\\=P(Z>\frac{1}{0.35)} \\=0.00214[/tex]
So option A is right answer.
