Question

If the variance of the poissons distribution is 2,find the probability for r=1,2,3 and 4 from the recurrence relation of the poisson distribution.Also find P(r≥4).​

Answer 1

SOLUTION

GIVEN

The variance of the poisson distribution is 2

TO DETERMINE

The probability for r = 1,2,3 and 4 from the recurrence relation of the poisson distribution.

Also find P(X ≥ 4)

EVALUATION

If X is a poisson variate then the probability density function is given by

$\displaystyle\sf{P(X = r) = {e}^{ - \mu} \frac{ { \mu}^{r} }{r !} } \: \: where \: r = 1,2,3,4..$

Which is required recurrence relation

Now it is given that variance = 2

$\sf{ \therefore \: \: \mu = 2 }$

$\displaystyle\sf{P(X =r ) = {e}^{ - 2} \: \frac{ {2}^{r} }{r !} }$

Thus

$\displaystyle\sf{P(X =1 ) = {e}^{ - 2} \: \frac{ {2}^{1} }{1 !} = 0.27 }$

$\displaystyle\sf{P(X =2 ) = {e}^{ - 2} \: \frac{ {2}^{2} }{2 !} = 0.27 }$

$\displaystyle\sf{P(X =3 ) = {e}^{ - 2} \: \frac{ {2}^{3} }{3 !} = 0.18 }$

$\displaystyle\sf{P(X =4) = {e}^{ - 2} \: \frac{ {2}^{4} }{4!} = 0.09}$

Now

$\displaystyle\sf{P(X \geqslant 4 ) }$

$\displaystyle\sf{ =1 - P(X < 4 ) }$

$\displaystyle\sf{ =1 - \bigg[ P(X =1 ) + P(X = 2) + P(X =3) \bigg]}$

$\displaystyle\sf{ =1 - \bigg[ 0.27 + 0.27 + 0.18 \bigg]}$

$\displaystyle\sf{ =1 - 0.72}$

$\displaystyle\sf{ =0.28}$

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