Answer:
(a) 0.2061
(b) 0.2514
(c) 0
Step-by-step explanation:
Let <em>X</em> denote the heights of women in the USA.
It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.
(a)
Compute the probability that the sample mean is greater than 63 inches as follows:
![P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z](https://tex.z-dn.net/?f=P%28%5Cbar%20X%3E63%29%3DP%28%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3E%5Cfrac%7B63-64%7D%7B3%2F%5Csqrt%7B6%7D%7D%29%5C%5C%5C%5C%3DP%28Z%3E-0.82%29%5C%5C%5C%5C%3DP%28Z%3C0.82%29%5C%5C%5C%5C%3D0.20611%5C%5C%5C%5C%5Capprox%200.2061)
Thus, the probability that the sample mean is greater than 63 inches is 0.2061.
(b)
Compute the probability that a randomly selected woman is taller than 66 inches as follows:
![P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28X%3E66%29%3DP%28%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cfrac%7B66-64%7D%7B3%7D%29%5C%5C%5C%5C%3DP%28Z%3E0.67%29%5C%5C%5C%5C%3D1-P%28Z%3C0.67%29%5C%5C%5C%5C%3D1-0.74857%5C%5C%5C%5C%3D0.25143%5C%5C%5C%5C%5Capprox%200.2514)
Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.
(c)
Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:
![P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0](https://tex.z-dn.net/?f=P%28%5Cbar%20X%3E66%29%3DP%28%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3E%5Cfrac%7B66-64%7D%7B3%2F%5Csqrt%7B100%7D%7D%29%5C%5C%5C%5C%3DP%28Z%3E6.67%29%5C%5C%5C%5C%5C%20%3D0)
Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.