# Distribution of $\frac{(Z_1 - \mu)^2}{\sigma^2}$

This is from an assignment.

The problem:

Given $n=15$ and assuming $Z_1, ..., Z_n$ are iid $N(\mu, \sigma^2)$, let $\bar{Z}$ be the sample mean and $S^2 = \frac{1}{n-1} \sum^n_{i=1} (Z_i - \bar{Z})^2$, define the following random variable $$V = \frac{(Z_1 - \mu)^2}{\sigma^2}$$ and calculate $$P(V > 1.801)$$

My solution:

The second part is not an issue, I am a little stuck on what distribution $V$ follows. I know that $(Z - \mu)$ subtracts the mean from the normal distribution and gives $\mu = 0$, and I know that $Z^2$ gives a $\chi^2$ distribution. I think that dividing by $\sigma^2$ probably standardizes the variance to produce a standard normal $\chi^2$ if that exists. To my mind though, that would look like $$\left( \frac{Z_1 - \mu}{\sigma^2} \right)^2$$ to first create the standard normal then square to produce a $\chi^2$ distribution.

Any prods in the right direction would be appreciated.

Edit: Some of the above information may be superfluous and for different parts of the question which I have not asked about.

• $X = (Z-\mu)/\sigma$ has a standard normal distribution and $V = X^2$. What is the distribution of the square of a standard normal? – linksys Feb 25 '16 at 17:37
• Dividing by the standard deviation causes the random variable to have unit variance. This is because constants are squared when they factor out of the variance operator. $V$ is just a standard normal squared. – dsaxton Feb 25 '16 at 17:38
• @linksys Is it a $\chi^2$ with 1 df? – Chris C Feb 25 '16 at 17:41
• @dsaxton Thanks! I didn't know that, and it makes sense as to why $\frac{\bar{Z} - \mu}{ \sigma / \sqrt{n}} \sim N(0, 1)$. A standard normal squared would be $\chi^2$ with 1 df? – Chris C Feb 25 '16 at 17:42
• Either there are a very large number of typos in this question or else most of it consists of intentional distractors. Since the definition of $V$ involves only $Z_1$, $\mu$, and $\sigma$, then--once you have double-checked that you copied it correctly!--you ought to completely ignore the information about $n$, $Z_2, \ldots, Z_n$, $S^2$, and $\bar Z$. – whuber Feb 25 '16 at 17:43

Dividing by the standard deviation causes a random variable to have unit variance. This is because constants are squared when they factor out of the variance operator. $V$ is just a standard normal squared which follows a $\chi^2_1$ distribution.