This question was previously asked in

DSSSB TGT Maths Female Subject Concerned - 18 Nov 2018 Shift 3

Option 4 : √2

__Concept:__

The **Probability function** for the Poisson distribution is defined as

\(P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}\) where \(\lambda \) is a parameter.

**Standard deviation **of X is \(\sqrt{\lambda }\).

__Calculation:__

We have 2P(X = 0) = P(X = 2). so, we should find these.

\(P(X=0)=\frac{e^{-\lambda}\lambda^0}{0!}=e^{-\lambda}\)

\(P(X=2)=\frac{e^{-\lambda}\lambda^2}{2!}=\frac{e^{-\lambda}\lambda^2}{2}\)

As we have,

2P(X = 0) = P(X = 2)

Putting in this we have,

\(2e^{-\lambda}=\frac{e^{-\lambda}\lambda^2}{2}\)

\(\lambda ^2=4\)

\(\lambda =\pm2\)

As, we can't take the negative value,

So, \(\lambda =2\)

So, the standard deviation of X =\(\sqrt{\lambda }\) = \(\sqrt2\).

**Hence, the Standard deviation of X is \(\sqrt2\)**