Probability that arithmetic mean > 171 cm, for a normal distribution

Question

I am wondering how to solve questions like this one:

There are a group of men are of heights which are normally distributed with μ = 173 cm and σ = 20 cm. A random sample of 300 men is chosen.

What is the probability that the arithmetic mean is greater than 171 cm?

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I am not familiar with how exactly to solve questions like these. I assume I need to use statistical tables as the final step to get the probability, but I don't know what to do before that. Am I supposed to calculate a z-value?

Answer 1

Let $S=X_1+\dots+X_n$, then $E(S)=n\cdot\mu$ and $V(S)=n\cdot\sigma^2$. As $\bar x=\frac1n S$ we have $E(\bar x)=\frac1n E(S)=\mu$ and $V(\bar x)=\frac{1}{n^2}V(S)=\sigma^2/n$.

In our case: Expected value of $\bar x=173$, standard deviation of $\bar x= 20/\sqrt{300}\approx1.155$, thus $P(\bar x>171)=1-\Phi\left(\dfrac{171-173}{1.155}\right)\approx1-0.042=0.958$.

Answer 2

The sample mean of Gaussian distribution $\mathcal{N}(\mu, \sigma^2)$ is again a Gaussian distribution $\mathcal{N}(\mu, \frac{\sigma^2}{N})$, where $N$ is the sample size. The get the probability that the arithmetic mean is greater than 171 cm you simply have to integrate this probability density function from 171cm to $\infty$. Maybe this might help you.