I was looking at an exercise which is related to the central limit theorem, but I don't understand a step of the solution, which is when it switches between calculating the probabilities that a sample mean is between an interval and calculating that the random variable $Z_n$ is between its corresponding interval. $Z_n$ should be defined as follows $$Z_n = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$$ In the case of the exercise, $\mu = 0$, $\sigma = \frac{1}{5}$ and the sample size is $n = 15$.
The exercise is
Take a random sample of size $n = 15$ from a distribution whose probability density function is: $$f(x)=\frac{3}{2}x^2$$ for $−1 < x < 1$. What is the probability that the sample mean falls between $\frac{−2}{5}$ and $\frac{1}{5}$?
And the solution first finds the expected value, which is $0$ and the standard deviation, which is $\frac{1}{5}$ and then it states the following $$P\left(\frac{-2}{5}<\overline{X}< \frac{1}{5} \right)=P(−2<Z<1)$$
But I don't understand that equation. Where does the $2$ and the $1$ come from from the left-side expression?
From my calculations we should have something like $$P\left( \frac{\frac{-2}{5} - 0}{\frac{\frac{1}{5}}{\sqrt{15}}} < Z_n < ...\right)$$ but $\frac{\frac{-2}{5} - 0}{\frac{\frac{1}{5}}{\sqrt{15}}}$ is not $-2$...
You can find the original problem at
The variance of the underlying distribution is $E(X^2)-(E(X))^2$, which is $3/5$. For $E(X^2)=\int_{-1}^1 \frac{3}{2}x^4\,dx$.
The variance of $\bar{X}$ is $3/5$ divided by $15$, which is $1/25$, so the standard deviation of $\bar{X}$ is $1/5$. (I believe you thought the standard deviation of the underlying distribution was $1/5$, in which case we would indeed have to adjust that to find the standard deviation of $\bar{X}$.)
Thus $-2/5$ is $2$ "standard deviation units" below the mean of $\bar{X}$, and $1/5$ is $1$ standard deviation unit above the mean, so the probability is indeed $\Pr(-2\lt Z\lt 1)$.