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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. $1$

$\hspace{2 cm}\underbrace{P(X\leq -1)}=P(Y\geq2)$
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 $\sigma ^{2}=9.$

edited
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open eyes, look what is given in the 1st image.
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Well explained. Thanks :-)
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why minus sign used in last line

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Hello set2018

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

So $P(Z≤-1)=P(Z≥1)$ // compare this with  $P(Z≤-1)$=$P(Z≥$ $\frac {3} {σ})$
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@Rupendra Choudhary
How we got this
-(-1) = 3/sigma(y)

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because $+1=-(-1)$
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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.

So,

$P\{Y\geq 2\}=P\{Y\gt 2\}$
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) $-(-1)=\dfrac{3}{\sigma}\Rightarrow \sigma = 3.$

edited
+1 vote
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yes it is.