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+29 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

- $3$
- $2$
- $\sqrt{2}$
- $1$

+36 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}$$

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

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

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

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

+6

Hello set2018

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

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

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

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

+31 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

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

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

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

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