Answer:
A. 0.0021
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 =
![(2.1)/(√(36) ) =0.35](https://img.qammunity.org/2020/formulas/mathematics/high-school/xp93xn1h1cqpnmsxcf6nvfkbcqm17758jb.png)
The sample entries X the heights are normal with mean= 66.9 inches and std deviation = 0.35 inches
Or we have
Z =
![(x-66.9)/(0.35)](https://img.qammunity.org/2020/formulas/mathematics/high-school/ve4sn1raibx4foscntvalqpyhfvn4opl1e.png)
Hence the probability that they have a mean height greater than 67.9 inches
=
![P(X>67.9)\\=P(Z>(1)/(0.35)) \\=0.00214](https://img.qammunity.org/2020/formulas/mathematics/high-school/whmldpmd1oj9rdx7p4z0wyo4zuzwsgje29.png)
So option A is right answer.