Answer is **A**

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

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 $\sigma ^{2}=9.$