Q81 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics

In a poisson distribution with unit mean, \(\sum_{x=0}^{\infty}|x-\bar{x}| P(X=x)=(\bar{x}\) is the mean of the distribution)

  • A
    e
  • B
    \(\frac{1}{\mathrm{e}}\)
  • Correct answer
    C
    \(\frac{2}{\mathrm{e}}\)
  • D
    \(\frac{2}{3 e}\)

Solution

For the unit mean, \(\bar{x}=1\) \(\begin{aligned} & \text {so } \sum_{x=0}^{\infty}|x-\bar{x}| P(X=x)=\sum_{x=0}^{\infty}|x-1| \frac{e^{-1}}{x !} \\ & =\frac{1}{e}\left[\frac{1}{0 !}+\sum_{x=1}^{\infty} \frac{x-1}{x !}\right]=\frac{1}{e}\left[1+\sum_{x=1}^{\infty} \frac{1}{(x-1) !}-\sum_{x=1}^{\infty} \frac{1}{x !}\right] \end{aligned}\) \(=\frac{1}{e}[1+e-(e-1)]=\frac{1}{e}(1+e-e+1)=\frac{2}{e}\) Hence, option (3) is correct.

Hence, the correct answer is option: 'C'.

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