I'm trying to solve the following question from the book All of Statistics page 83:
"Suppose that the height of men has mean 68 inches and standard deviation 2.6 inches. We draw 100 men at random. Find (approximately) the probability that the average height of men in our sample will be at least 68 inches."
My solution:
Define $ \overline{X}_{100} = \dfrac{\sum_{i=1}^{100}X_{i}}{100} $. By CLT, we have
$P(\overline{X}_{100} \geq 68) = 1-P(\overline{X}_{100} \leq 68) = 1- P(\dfrac{\sqrt{100}(\overline{X}_{100}-68)}{2.6} \leq 0) = 1 - \Phi(0)= \frac{1}{2}$.
Now, I have two questions:
Is my solution true?
If the answer to question 1 is yes, what's the intuition? I mean, now, because we are working at point 0, (according to my solution) the answer is independent of the values of $ \sigma $ and $ n = 100 $. If my solution is wrong, can anyone correct it?
Regards,
It's correct. The intuition is simply that the normal distribution is symmetric. If you subtract each height from 138, that's still normally distributed with mean 68 and standard deviation 2.6. So the probability that the average of (136-height) in a sample is at least 68 is the same as the probability that the average height is at least 68, which in turn is the same as the probability that the average height is more than 68 (since it has probability $0$ of being exactly 68). But these are complementary events, so this probability is $1/2$.
This argument works irrespective of the sample size and standard deviation.