Question

Let X be a normal random variable with mean 1 and variance 4. The probability P(X < 0) is

• A. 0.5
• B. greater than zero and less than 0.5
• C. greater than 0.5 and less than 1.0
• D. 1.0

Answer 1

The correct option is B greater than zero and less than 0.5

Given, a normal random variable, X with

$\text{mean}\left(\overline{X}\right)=1$

$\text{variance}\left({\sigma }^2\right)=4$

$\text{Standard deviation}=\sqrt{{\sigma }^2}=\sqrt{4}=2$

We know that probability density function of normal random variable is

$P\left(X\right)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{1}{2}{\left(\frac{X-\overline{X}}{\sigma }\right)}^2}$

Putting the value of $\overline{X}\text{and}\sigma$ we get

$P\left(X\right)=\frac{1}{2\sqrt{2\pi }}{e}^{-\frac{(X-1{)}^2}{8}}$

at X = 0

$P\left(0\right)=\frac{1}{2\sqrt{2\pi }}{e}^{-\frac{-1}{8}}$

$P\left(0\right)=0.176$

at X = -1

$P(-1)=\frac{1}{2\sqrt{2\pi }}{e}^{-\frac{(-1-1{)}^2}{8}}=0.1209<0.5$

Hence increasing the value of X in negative direction, P(X) will decrease and tend to zero.

$\therefore$ P(X < 0) always lies in the range 0 < P(X < 0) < 0.5.

Alternative Solution

$\text{Mean}\left(\overline{X}\right)=1,\sigma =2$

So, $Z_x=\frac{x-\overline{x}}{\sigma }=\frac{0-1}{2}=-\frac{1}{2}$

So, $P(x<0)=P(Zx<-\frac{1}{2})=P(Z_x>\frac{1}{2})$

$=\frac{1}{2}-P(0\le Z_x\le \frac{1}{2})<0.5$