Question

If for a Poisson distribution $P(X = 0) = 0.2$ , then the variance of the distribution is
A. 5
B. ${\log _{10}}5$
C. ${\log _e}5$
D. ${\log _5}e$

Answer 1

Since in Poisson distribution, the mean and variance are equal, we need to find $\lambda$. So, we will use the formula in poisons theorem to find the value of $\lambda$. we would need to take a logarithm on both sides of our equation in order to eliminate ‘e’ and solve the equation to get to the final answer.

Complete step by step solution:
According to the question given
$P(X = 0) = 0.2$ … (1)

We know that, according to Poisson's theory
$P(X = x) = \dfrac{{{\lambda ^x} \times {e^{ - \lambda }}}}{{x!}}$ … (2)
Where $\lambda$ is the mean and variance of the given distribution

So, when $x = 0$ , we get
$\Rightarrow P(X = 0) = \dfrac{{{\lambda ^0} \times {e^{ - \lambda }}}}{{0!}}$ … (3)

Comparing (1) and (3), we get
$\Rightarrow 0.2 = \dfrac{{{\lambda ^0} \times {e^{ - \lambda }}}}{{0!}}$

We know that ${\lambda ^0} = 1$ and $0! = 1$ , hence
$\Rightarrow 0.2 = \dfrac{{1 \times {e^{ - \lambda }}}}{1}$
On simplification we get,
$\Rightarrow 0.2 = {e^{ - \lambda }}$

Taking logarithm both side, we get
$\Rightarrow {\log _e}0.2 = {\log _e}{e^{ - \lambda }}$ … (4)

We know that,
$\bullet {\log _e}{a^b} = b{\log _e}a$
$\bullet - {\log _e}a = {\log _e}\dfrac{1}{a}$
$\bullet {\log _a}a = 1$

Hence from (4), now use the above formulas to simplify, we get
$\Rightarrow {\log _e}0.2 = - \lambda \times {\log _e}e$

Using, ${\log _a}a = 1$ we get
$\Rightarrow {\log _e}0.2 = - \lambda$
On multiplying by -1 on both the sides we get,
$\Rightarrow \lambda = - {\log _e}0.2$

Using $- {\log _e}a = {\log _e}\dfrac{1}{a}$ , we get
$\Rightarrow \lambda = {\log _e}\dfrac{1}{{0.2}}$
Hence we have,
$\Rightarrow \lambda = {\log _e}5$

Since $\lambda$ is the variance of the given distribution

Hence, the final answer is C.

Note:
In Poisson's distribution, the mean and variance are the same and given by. So, the question may ask us about mean, or it can ask us about the variance, the steps to be followed remains unaffected. Whenever we come across ‘e’ in any equation, there are high chances that we need to use logarithm at both sides of the equation because it helps us eliminate ‘e’ from the equation, elimination of ‘e’ instantly makes working on that equation very easy.