Question

Let $p(x)$ represent the probability mass function of a Poisson distribution. If its mean $\lambda=3.725$, then value of $x$ at which $p ( x )$ is maximum is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

In Poisson distribution
$p(x)=\frac{e^{-\lambda \lambda^{r}}}{r !} $
$\Rightarrow p(2)=\frac{e^{-\lambda}(3.725)^{2}}{2 !}$
$[\because \lambda=3.725]$
$p(2)=e^{-\lambda}\left(\frac{13.875}{2}\right)=e^{-\lambda}(6.937) $
$p(3)=\frac{e^{-\lambda}(3.725)^{3}}{3 !}=e^{-\lambda}(8.614) $
$p(4)=\frac{e^{-\lambda}(3.725)^{4}}{4 !}=e^{-\lambda}(8.022)$
$p(5)=\frac{e^{-\lambda}(3.725)^{5}}{5 !}=e^{-\lambda}(5.976)$
Clearly, $p(3)$ has maximum value
$\therefore $ Value of $x=3$