Question

A set of normally distributed data has a mean of 485 and a standard deviation of 11.6. Find the probability of randomly selecting 40 values and getting a mean less than 490.

Answer 1

Let $X_i$ denote a data point taken from the distribution, where $1\le i\le40$, and let $Y$ denote the average.

You want to find

$\mathbb P\left(\displaystyle\frac1{40}\sum_{i=1}^{40}X_i$

First, let's recall a few things. The PDF of a normal distribution with mean $\mu$ and variance $\sigma^2$ is

$f(x;\mu,\sigma^2)=\displaystyle\frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$

Each of the $X_i$ are presumably independently selected, so they are i.i.d. random variables.

The MGF of a normal distribution is

$M_X(t)=\mathbb E(e^{tX})$
$M_X(t)=\displaystyle\int_{-\infty}^\infty e^{tx}f_X(x)\,\mathrm dx$
$M_X(t)=\exp\left(\mu t+\dfrac12\sigma^2t^2\right)$

The MGF of a linear combination of i.i.d. random variables is

$M_{c_1X_1+\cdots+c_nX_n}=M_{X_1}(c_1t)\times\cdots\times M_{X_n}(c_nt)=\displaystyle\prod_{i=1}^nM_{X_i}(c_it)$

In this case, each $c_i=\dfrac1{40}$. This product of MGFs reduces to an MGF of a normal distribution because the $X_i$ are i.i.d..

$M_Y(t)=\displaystyle\prod_{i=1}^{40}M_{X_i}(t)=\exp\left(\mu\left(\dfrac t{40}\right)+\frac12\sigma^2\left(\dfrac t{40}\right)^2\right)\times\cdots\times\exp\left(\mu\left(\dfrac t{40}\right)+\frac12\sigma^2\left(\dfrac t{40}\right)^2\right)$
$M_Y(t)=\exp\left(\mu t+\dfrac12\left(\dfrac\sigma{\sqrt{40}}\right)^2t^2\right)$

which is indeed the MGF of a normal distribution with mean $\displaystyle\mu\sum_{i=1}^{40}\frac1{40}=\mu$ and variance $\sigma^2\displaystyle\sum_{i=1}^{40}\left(\frac1{40}\right)^2=\frac{\sigma^2}{40}$

So, the PDF of $Y$, given that $\mu=485$ and $\sigma=11.6$, is

$f_Y(y)=\displaystyle\frac1{\frac{11.6}{\sqrt{40}}\sqrt{2\pi}}\exp\left(-\frac{(y-485)^2}{2\left(\frac{11.6}{\sqrt{40}}\right)^2}\right)$

Now,

$\mathbb P(Y$

or, using the CDF of $Y$,

$\mathbb P(Y$