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?

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?


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 $P(\bar x>171)=1-\Phi\left(\dfrac{171-173}{1.155}\right)\approx1-0.042=0.958$.

  • $\begingroup$ How do I know when to use N(μ,σ²/N) instead of just N(μ,σ²)? $\endgroup$ – mangopancake Nov 27 '13 at 21:23
  • $\begingroup$ @Laura I've extended my answer. $\endgroup$ – Michael Hoppe Nov 28 '13 at 6:44

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.

  • $\begingroup$ If I integrate the probability I get is 0.5398, but would that not be the probability that a man in the group is taller than 171 cm, as oppose to the arithmetic mean being greater than 171 cm? I'm not sure how to think about probability of the arithmetic mean being greater than a certain number, because isn't the mean already determined as 173 cm? $\endgroup$ – mangopancake Nov 26 '13 at 21:55
  • $\begingroup$ The sample mean is distributed according to $\mathcal{N}(173 cm, \frac{20}{300} cm)$, i.e., the standard deviation is 0.6 mm. If you integrate this Gaussian from 171 cm to $\infty$, you should get a much larger probability, since it is very unlikely that the sample mean is that low or lower for such a large sample. $\endgroup$ – bluenote10 Nov 26 '13 at 22:05
  • $\begingroup$ $\bar x$ is $N(\mu,\sigma/\sqrt{n})$-distributed. $\endgroup$ – Michael Hoppe Nov 27 '13 at 10:07
  • 1
    $\begingroup$ You are writing here $N\left(\mu,\sigma\right)$ and $N\left(\mu,\frac{\sigma}{N}\right)$. That should be $N\left(\mu,\sigma^{2}\right)$ and $N\left(\mu,\frac{\sigma^{2}}{N}\right)$. Probably nothing more than a typo, but a dangerous one: notice that we do not have a deviation of $\frac{\sigma}{N}$ but one of $\frac{\sigma}{\sqrt{N}}$. $\endgroup$ – drhab Nov 27 '13 at 10:23
  • $\begingroup$ @drhab: Yes, definitely a dangerous typo: The variance is $\frac{(20 cm)^2}{300}$ and the standard deviation is $\frac{20 cm}{300} = 1.155 cm$. However, still a high probability that $\bar x$ is larger than $171 cm$. $\endgroup$ – bluenote10 Nov 27 '13 at 11:05

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