Question: Suppose that a random variable $T$ is equal to $\frac{Z}{\sqrt{U/25}}$, where $Z$ is N(0,1) and $U$ is $\chi^2(25)$. What is the probability that this random variable takes a value between 1.708 and 2.787?
Approach: So $T = \frac{\overline{\chi}-\mu}{S/\sqrt{n}}$ where: $\mu = 0$, $n = 25$ and $S = 1$.Then, the probability to find $T$ between the specified values is:
$P(1.78 < T < 2.787)$ $\rightarrow$ $P(\frac{1.708-0}{1/\sqrt{25}} < T < \frac{2.787-0}{1/\sqrt{25}})$ $\rightarrow$ $P(8.54 < T < 13.935)$.
I am unsure on how to proceed or even if what I did is correct.
You can conclude the distribution of T from sampling. Lets say you take a random sample of size $n$ from a normal distribution. Therefore you have $n$ (independent) random variables ($X_i$) with the distribution $\mathcal N(\mu, \sigma^2)$
The unbiased estimator of the variance is $s_u^2=\frac{1}{n-1}\sum_{i=1}^n (X_i-\mu)^2$. Now we can divide both sides of the equation by $\sigma^2$
$Q=\frac{s_u^2}{\sigma^2}=\frac{1}{n-1}\sum_{i=1}^n \left(\frac{X_i-\mu}{\sigma}\right)^2$
This is equal to $Q=\frac{1}{n-1}\sum_{i=1}^n Z_i^2$, where $Z_i\sim \mathcal N(0,1)$
From the sample theory it is known that $s=(n-1)Q\sim \chi^2_{n-1}$
And finally we know that $\frac{Z}{s/ \sqrt n}$ is $t$-distributed as $t_{n-1}$
Hint: The result in percentage has only one decimal place.