## Question

Download Solution PDF# If X is a Poisson random variate with mean 3, then P{|X - 3| < 1} will be

## Answer (Detailed Solution Below)

## Detailed Solution

Download Solution PDF**Concept:**

Poisson distribution formula,

\(\;P\left( x \right) = \frac{{{e^{ - λ }}{λ ^x}}}{{x!}}\)

where λ = mean value of occurrence within an interval

P(x) = probability of x occurrence within an interval

**Calculation:**

**Given:**

Mean (λ) = 3, x = 3

\(\;P\left( x \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}\)

\(\;P\left( x=3 \right) = \frac{{{e^{ - 3 }}{3 ^3}}}{{3!}}\)

\(\;P\left( x=3 \right) = \frac{{{e^{ - 3 }}{(3\times3\times3)}}}{{3\times2\times1}}\)

**\(\;P\left( x=3 \right) = \frac{{{e^{ - 3 }}{(9)}}}{{2}}\)**

**P(|x - 3| < 1) = P(x = 3)**

**because for only x = 3, (|x - 3| < 1) is possible, for any other value of x, |x - 3| will become greater than 1.**

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