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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?

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    $\begingroup$ Because the world cannot seem to settle on a standard... In "$\mathcal{N}(1,4)$", is the "$4$" the variance or the standard deviation? $\endgroup$ – Eric Towers Mar 17 '16 at 22:12
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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 $$P(Y<-1\cup Y>1) = P(|Y|>1) = 1-P(|Y|< 1) = 1-P(-1<Y<1).$$

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$Y$ 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?

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    $\begingroup$ Why is $X \sim N(1,4)$, the same as all the $X_i$? $\endgroup$ – Dilip Sarwate Mar 17 '16 at 22:18
  • $\begingroup$ Oh I lost a denominator, thank you! $\endgroup$ – iiivooo Mar 17 '16 at 22:24
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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 $0.769411\dots$.

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  • $\begingroup$ The numerical value is approximately 0.8653. $\endgroup$ – Sungjin Kim Mar 18 '16 at 0:14
  • $\begingroup$ So it is... I wonder if that's left over from running the process assuming $4$ was the standard deviation. $\endgroup$ – Eric Towers Mar 18 '16 at 1:13
  • $\begingroup$ @i707107 : Turns out it 0.769411... was left over from when I assumed "$4$" was the standard deviation. Corrected, extended, and labeled. $\endgroup$ – Eric Towers Mar 18 '16 at 1:21

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