edited by
23,472 views
52 votes
52 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. $1$
edited by

3 Answers

Best answer
61 votes
61 votes

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
38 votes
38 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
Answer:

Related questions

1 votes
1 votes
0 answers
1
0 votes
0 votes
2 answers
2
Pooja Khatri asked Sep 26, 2018
1,178 views
Assume that $X$ is Normal with mean $\mu$ $=$ $2$ and variance $\sigma^2$ $=$ $25$. Compute the probability that $X$ is between $1$ and $4$.
0 votes
0 votes
2 answers
3
Pooja Khatri asked Sep 26, 2018
620 views
What is the probability that a Normal random variable differs from its mean $\mu$ by more than $\sigma$ ?
4 votes
4 votes
2 answers
4
Pooja Khatri asked Sep 26, 2018
730 views
Let X be a $N(\mu , \sigma^2)$ random variable and let $Y = \alpha X+\beta$, with $\alpha$ $0$. How is $Y$ distributed?