How to evaluate cumulative distribution function w/ sigma of given SRS? I have been given a set of 10 positive values that make up an SRS (Simple Random Sample) of a population, which has a normal distribution.
I then am asked to evaluate $a = \sum_{i=1}^n\frac{(x_{i}-\mu)^2}{\sigma^2}$ and provide the distribution. Since it is a sum, I assumed a would simply be a number.
Finally, I am asked to evaluate the following CDF: P ($\sum_{i=1}^n\frac{(x_{i}-\mu)^2}{\sigma^2} \le a$).
How should I interpret this? If the values in the SRS are all positive, then how could any subset less than the size of $n$ have the value of $a$? Wouldn't the probability of this just be $1$?
 A: Since $\dfrac{x_i-\mu}{\sigma}\sim\operatorname N(0,1)$ and $x_1,\ldots,x_n$ are mutually independent, we have $ \displaystyle \sum_{i=1}^n \left( \frac{x_i-\mu} \sigma \right)^2 \sim \chi^2_n.$
A: For many years, there have printed CDF tables of the chi-squared distribution in most applications oriented basic statistics texts.
Also, statistical software is now widely used: For example, Suppose $Q \sim\mathsf{Chisq}(\nu = 5),$ and that you want $P(Q \le .05) = 0.0000292$ or you want $c=11.0705$ such that
$P(Q \le c) = .95,$ you can use the chi-squared CDF pchisq or the chi-squared quantile function (inverse CDF) qchisq in R, as follows:
pchisq(.05, 5)
[1] 2.920954e-05
qchisq(.95, 5)
[1] 11.0705
qchisq(.05, 5)
[1] 1.145476       # P(Q < 1.1455) = 0.05
pchisq(1.1455, 5)
[1] 0.05000219

Also, the chi-squared density function is dchisq.

hdr = "Density of CHISQ(df = 5)"
curve(dchisq(x, 5), 0, 15, lwd=2, col="blue", 
  ylab="PDF", xlab="q", main=hdr)
 abline(h=0, col="green2"); abline(v=0, col="green2")
 abline(v = c(.05, 1.1455, 11.0705), col=c("red","cyan3", "purple"), lwd=2, lty="dotted")

Application: Suppose you want a 90% confidence interval
for $\sigma^2$ based on a sample of size $n=10$ from
a normal population with unknown mean and variance.
Then $\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu = 9).$ [The degrees of freedom are reduced to $\nu=n-1=9$ because
$\bar X$ estimates $\mu$ in the usual formula for the
sample variance $S^2.$ Easy to remember; a bit messy to prove.]
Then if values $L$ and $U$ cut
5% of the probability from the lower and upper tails, respectively of $\mathsf{Chisq}(9),$ we have
$P\left(L \le \frac{9S^2}{\sigma^2} \le U\right) = 0.9.$ This
leads to a 90% CI for $\sigma^2$ of the form
$\left(\frac{(n-1)S^2}{U}, \frac{(n-1)S^2}{L}\right).$
In particular, suppose we have a normal sample of size $n=10$ with sample variance $S^2 = 3.21,$ then our 90% CI for $\sigma^2$ is
$(1.71, 8.69).$ [Generally, it takes a lot of data to
get a short CI for a population variance. "Variances are very variable."]
9*3.21/qchisq(c(.95,.05), 9)
[1] 1.707550 8.688427

