Question

statistics question:

The percentage changes in annual earnings for a company are normally distributed with a mean of 5% and a standard deviation of 12%. The probability that the average change in earnings over the next five years will be greater that 15.5% is?

Answer 1

I know I have to normalize and check the area under the normal distribution table so \[\frac{15.5-5}{standard deviation}\] but apparently I must divide the standard deviation by the square root of five... why?

Answer 2

The percent changes themselves are normally distributed, but you are asked for a probability regarding the *average* change.

So if percent change can be modeled by a normal random variable \(X\), then the *average* change is modeled by \(\overline{X}=\dfrac{1}{n}\sum\limits_{k=1}^nX_k\).

You can show that if \(X\) has standard deviation \(\sigma\), then \(\overline{X}\) has standard deviation \(\dfrac{\sigma}{\sqrt n}\).

Answer 3

Here's the derivation. Assume for \(k=1,2,\ldots,n\) that \(X_k\) are independent and identically distributed, each with mean \(\mu\) and standard deviation \(\sigma\).
\[\begin{align*}
\overline{X}&=\frac{1}{n}\sum_{k=1}^n X_k\\\\
\implies E(\overline{X})&=E\left(\frac{1}{n}\sum_{k=1}^n X_k\right)\\
&=\frac{1}{n}E\left(\sum_{k=1}^n X_k\right)\\
&=\frac{1}{n}\sum_{k=1}^nE(X_k)\\
&=\frac{n}{n}E(X_k)\\
&=E(X_k)\\
&=\mu\\\\
\implies V(\overline{X})&=V\left(\frac{1}{n}\sum_{k=1}^n X_k\right)\\
&=\frac{1}{n^2}\sum_{k=1}^nV(X_k)\\
&=\frac{n}{n^2}\sigma^2\\
\sqrt{V(\overline{X})}&=\frac{\sigma}{\sqrt n}
\end{align*}\]