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 $x=3$