Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is $P(|Y| > 1)$? 
Suppose that $X_1,X_2,\ldots,X_{25}$ are independent random variables from $\mathcal{N}(1, 4)$. Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is the probability that $|Y| > 1$? 

We know that $Y$ is a standard normal random variable.
I keep trying to transform it into normal distribution, but I keep getting the wrong answer. The answer is $0.6826$. How do I solve this?
 A: $Y$ is not a standard normal random variable!
If $X_i \sim N(1,4)$, then $X=\frac{1}{25}\sum_{i=1}^{25} X_i \sim N(1,\frac{4}{25}).$ So $Y$ is normal with mean 
$$EY=\frac{5}{2} -\frac{2}{5}=\frac{21}{10},$$ 
and  variance 
$$\textrm{var}(Y)=\left(\frac{5}{2}\right)^2 \cdot  \frac{4}{25} =1.$$
Can you finish it?  
A: Recall that for two iid normal variables $\mathsf X$, $\mathsf Y$ with mean $\mu$ and variance $\sigma^2$, then
$$\mathsf X+\mathsf Y\sim \mathcal N(2\mu, 2\sigma^2).$$
Further, for some constants $a$ and $b$ not $0$, we have
$$a\mathsf X+b\sim \mathcal N(a\mu+b, a^2\sigma^2).$$
Generalizing to the $n$ case, and assuming $\sigma^2  = 4$, we have that for our $X_1+\dotsb+X_{25}$,
$$X_1+\dotsb+X_{25}\sim \mathcal N(25, 100)$$
then 
$$X = \frac{1}{25}(X_1+\dotsb+X_{25})\sim\mathcal N(1,4/25).$$
Further, 
$$Y = \frac{5}{2}X-\frac{2}{5}\sim\mathcal N(21/10, 1)$$
Lastly, notice that
$$P(Y<-1\cup Y>1) = P(|Y|>1) = 1-P(|Y|< 1) = 1-P(-1<Y<1).$$ 
A: Assuming "$4$" is the variance, ...
$X$ is a linear function of i.i.d. r.v.s.  It's mean is $\frac{1}{25} ( 25 \cdot 1) = 1$.  It's variance is $\left( \frac{1}{25} \right)^2 (25 \cdot 4) = 4/25$.  (This is a standard result.  There are several ways to get this.  Google "sums of normally distributed variables" for a plethora.)
$Y$ is a linear function of $X$.  It's mean is $\frac{5}{2} \cdot 1 - \frac{2}{5} = \frac{21}{10}$.  It's variance is $\left( \frac{5}{2} \right)^2 \cdot \frac{4}{25} = 1$.  (This is another standard result.  Ask Professor Google for "linear transform of normally distributed variable".)
Note that $Y - \frac{21}{10}$ is a $\mathcal{N}(0,1)$ distributed random variable, so we can use standard tables to look up \begin{align*}
    |Y| &> 1  \\
    Y < -1 &\text{ or } Y > 1  \\
    Y - \frac{21}{10} < \frac{-31}{10} &\text{ or } Y - \frac{21}{10} > \frac{-11}{10}  \text{.}
\end{align*}  This is $\frac{1}{2} \left(\text{erf}\left(\frac{11}{10
   \sqrt{2}}\right)-\text{erf}\left(\frac{31}{10 \sqrt{2}}\right)\right) + 1 = 0.865302\dots$.
If, on the other hand, "$4$" is the standard deviation, the same process leads to the probability being $0.769411\dots$.
