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+22 votes

Let $X$ be a random variable following normal distribution with mean $+1$ and variance $4$. Let $Y$ be another normal variable with mean $-1$ and variance unknown. If $P (X ≤ -1) = P (Y ≥ 2)$ , the standard deviation of $Y$ is

  1. $3$
  2. $2$
  3. $\sqrt{2}$
  4. $1$
asked in Probability by Veteran (59.9k points) | 4k views

3 Answers

+25 votes
Best answer

Answer is A

$\hspace{2 cm}\underbrace{P(X\leq -1)}=P(Y\geq2)$
We can compare their values using standard normal distributions.

$\hspace{3 cm}X \hspace{3 cm} Y$
$\hspace{2 cm}\mu_{X}=+1\hspace{2 cm} \mu_{Y}=-1$
$\hspace{2 cm}{\large\sigma}_{X}{^2}=4\hspace{2 cm}{\large\sigma}_{Y}{^2}=?$

$\hspace{1 cm}Z_{X}=\dfrac{X-1}{\sqrt{4}}\hspace{1 cm}Z_{Y}=\dfrac{Y-(-1)}{{\large\sigma}_{Y}}$

$\hspace{1 cm}2Z_{X}+1=X\hspace{1 cm}Y={\large\sigma}_{Y}Z_{Y}-1$

$\quad P(2Z_{X}+1\leq -1)=P({\large\sigma}_{Y}Z_{Y}-1\geq 2)$

$\hspace{1 cm}\;\;\;\; P(Z_{X}\leq -1)=P(Z_{Y}\geq \dfrac{3}{{\large\sigma}_{Y}})$

$\hspace{2.8 cm} -(-1)=\dfrac{3}{{\large\sigma}_{Y}}$

$\hspace{3.5 cm}{\large\sigma}_{Y}=3$

The random variables $X$ and $Y$ can be written as $X=2U+1$ and $Y=\sigma V-1$
where $U$ and $V$ both have standard normal distribution.

$P\{U\geq 1\}=P\{U\leq -1\}=P\{X\leq -1\}=P\{Y\geq 2\}=P\left\{V\geq \dfrac{3}{\sigma}\right\}=P\left\{U\geq \dfrac{3}{\sigma}\right\}$

The first equality is a consequence of the fact that the standard normal distribution is symmetric.
The last equality is a consequence of the fact that $U$ and $V$ have the same distribution.

So $1=\dfrac{3}{\sigma}$, so that $\sigma ^{2}=9.$


answered by Boss (30.9k points)
edited by
open eyes, look what is given in the 1st image.
Well explained. Thanks :-)

why minus sign used in last line 


Hello set2018

in standard normal distribution $P(Z≤-a)=P(Z≥a)$

So $P(Z≤-1)=P(Z≥1)$ // compare this with  $P(Z≤-1)$=$P(Z≥$ $\frac {3} {σ})$

@Rupendra Choudhary
How we got this 
-(-1) = 3/sigma(y)

because $+1=-(-1)$
Also, for any continous random variable X

$P\{X=a\}=\int_a^{a}f(x)dx=0$ means the probability that a continous random variable will assume a particular value is 0.


$P\{Y\geq 2\}=P\{Y\gt 2\}$
Please suggest ,   where to read about theoritical point about it?
Before attending any question, first read the concept :)
Ok.. they are for standard normal distribution.( actually i read only basic concepts of Normal distribution). But it is clear now thanks.
+22 votes
First lets convert both $X$ and $Y$ to Standard normal distribution.



Now replace $X$ and $Y$ in $P(X\leq {-1})=P(Y\leq 2)$ we get $P(Z\leq {-1})=P\left(Z\geq \dfrac{3}{\sigma}\right)$

Since the Standard Normal Curve is symmetric about the mean( i.e, zero) $-(-1)=\dfrac{3}{\sigma}\Rightarrow \sigma = 3.$

Answer is Option A
answered by Active (2.1k points)
edited by
+1 vote
(A) is the correct answer?
answered by (33 points)
yes it is.

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