Let p(x) represent the probability mass funct | vedclass

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(B) The probability mass function of a Poisson distribution is given by $p(x) = \frac{e^{-\lambda} \lambda^x}{x!}$ for $x = 0, 1, 2, \dots$. To find t

Vedclass · Accepted Answer

$3$

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(B) The probability mass function of a Poisson distribution is given by $p(x) = \frac{e^{-\lambda} \lambda^x}{x!}$ for $x = 0, 1, 2, \dots$. To find the value of $x$ for which $p(x)$ is maximum, we examine the ratio $\frac{p(x)}{p(x-1)}$. $\frac{p(x)}{p(x-1)} = \frac{e^{-\lambda} \lambda^x / x!}{e^{-\lambda} \lambda^{x-1} / (x-1)!} = \frac{\lambda}{x}$. For $p(x)$ to be maximum, we need $\frac{p(x)}{p(x-1)} \geq 1$ and $\frac{p(x+1)}{p(x)} \leq 1$. This implies $\frac{\lambda}{x} \geq 1 \implies x \leq \lambda$ and $\frac{\lambda}{x+1} \leq 1 \implies x+1 \geq \lambda$. Thus, the mode $x$ satisfies $\lambda - 1 \leq x \leq \lambda$. Given $\lambda = 3.725$, we have $3.725 - 1 \leq x \leq 3.725$, which means $2.725 \leq x \leq 3.725$. Since $x$ must be an integer, the value of $x$ that maximizes $p(x)$ is $3$.

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

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

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