in Probability edited by
16,897 views
41 votes
41 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 $P (X \leq -1) = P (Y \geq 2)$ , the standard deviation of $Y$ is

  1. $3$
  2. $2$
  3. $\sqrt{2}$
  4. $1$
in Probability edited by
16.9k views

3 Answers

51 votes
51 votes
Best answer

Answer is A

$P(X\leq -1)=P(Y\geq2)$
We can compare their values using standard normal distributions:

$$\begin{array}{c|c}
X& Y\\
\mu_X = +1 & \mu_Y = -1\\
{\large\sigma}_{X}{^2}=4&{\large\sigma}_{Y}{^2}=?\\
Z_{X}=\dfrac{X-1}{\sqrt{4}}&Z_{Y}=\dfrac{Y-(-1)}{{\large\sigma}_{Y}}\\
2Z_{X}+1=X&Y={\large\sigma}_{Y}Z_{Y}-1\\
\end{array}$$

$\implies  P(2Z_{X}+1\leq -1)=P({\large\sigma}_{Y}Z_{Y}-1\geq 2)$
$\implies P(Z_{X}\leq -1)=P(Z_{Y}\geq \dfrac{3}{{\large\sigma}_{Y}})$

$\implies -(-1)=\dfrac{3}{{\large\sigma}_{Y}}$

$\implies {\large\sigma}_{Y}=3$

edited by

4 Comments

Because it is symmetric, that means the unknown number will be on the right side of zero but of same magnitude.

0
0
edited ago by

@manoj_mj thanks. Clear now. 

0
0

 copy the name & paste in ur reply

0
0
33 votes
33 votes
First lets convert both $X$ and $Y$ to Standard normal distribution.

$Z=\dfrac{X-1}{2}$

$Z=\dfrac{Y+1}{\sigma}$

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
edited by
4 votes
4 votes

this video might help.

by

1 comment

Thanks for sharing
1
1