Variance of Random Variable and Normal Variable

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 $$P(X\leq-1)=P(Y\geq2)$$ then standard deviation of Y is

My Solution:

$E(X)=\mu=1, V(X)=\sigma^2=4$ as

$\sigma^2=E[X^2]-(E[X])^2=E[X^2]-\mu^2$

$4=E[X^2]-1^2$

$\implies E[X^2]=5$

Now How can we calculate $\sigma^2$ for Y

• Your "solution" does not use the given equality. So - as to be expected - is not fruitful. – drhab Jan 23 '15 at 12:22

We will use the fact that $P(Z\le-1)=P(Z\ge1)$, where $Z$ is a standard normal random variable. We have that $$P(X\le-1)=P\biggl(\frac{X-1}2\le-1\biggr),$$ $$P(Y\ge2)=P\biggl(\frac{Y+1}\sigma\ge\frac3\sigma\biggr)$$ and $$P\biggl(\frac{X-1}2\le-1\biggr)=P\biggl(\frac{Y+1}\sigma\ge\frac3\sigma\biggr).$$ $(X-1)/2$ and $(Y+1)/\sigma$ are both standard normal random variables. Hence, $3/\sigma$ must be equal to $1$ and $\sigma$ to $3$.
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.
$P\{U\geq 1\}=P\{U\leq -1\}=P\{X\leq -1\}=P\{Y\geq2\}=P\{V\geq\frac{3}\sigma\}=P\{U\geq\frac{3}\sigma\}$
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 $1=\frac{3}\sigma$, so that $\sigma^2=9$.