# Using the (strong) law of large numbers problem

this is a practice problem I am doing. It says "Suppose that in a community the distributions of heights of men and women in centimeters are N(173, 40) and N(160, 20) respectively. Calculate P that the average height of 10 randomly selected men is at least 5 centimeters larger than the average height of six randomly selected women." I am tempted to say that I should just look at the probability that height of 1 randomly selected man is at least 5 cm greater than height of 1 randomly selected woman, but I am struggling to make this rigorous. Help appreciated! P.S. Central Limit Theorem might be used here too, this is a chapter review question so not sure what exactly can be useful.

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Consider the fact that from a normal distribution $N(\mu, \sigma)$, the distribution of means from samples of size $n$ is $N(\mu, \sigma/\sqrt{n})$. We can use this to make new distributions for the mean height of 10 random men, and the mean height of 6 random women; then consider the fact that the distribution for a random sample from $N(\mu_1, \sigma_1)$ minus a random sample from $N(\mu_2, \sigma_2)$ is $N(\mu_1-\mu_2, \sqrt{\sigma_1^2 + \sigma_2^2})$.

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I got to that last step, from there, do I just find the probability that $x \geq 5?$ Using normal distribution z tables etc. – prboq Nov 18 '11 at 6:15
Precisely!  – smackcrane Nov 18 '11 at 6:16
Thanks a lot, it's that last step I wasn't getting... – prboq Nov 18 '11 at 6:16

Hints: (1) What is the distribution of the average height of 10 randomly selected men? (2) What is the distribution of the average height of 6 randomly selected women? (3) What is the distribution of the difference of these two average heights?

There is no need of CLT nor of LLN here, rather one uses some properties of independent gaussian random variables.

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