The number of persons joining a cinema ticket | vedclass

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Question

(C) The Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$,where $\lambda = 6$.We need to find $P(1 \leq X \leq 5) = P(X=1)

Vedclass · Accepted Answer

$6 	imes e^{-6}(29.8)$

Vedclass · Answer

(C) The Poisson distribution is given by $P(X=x) = \frac{e^{-\lambda} \lambda^x}{x!}$,where $\lambda = 6$.We need to find $P(1 \leq X \leq 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)$.$P(1 \leq X \leq 5) = e^{-6} \left[ \frac{6^1}{1!} + \frac{6^2}{2!} + \frac{6^3}{3!} + \frac{6^4}{4!} + \frac{6^5}{5!} ight]$$= e^{-6} \left[ 6 + \frac{36}{2} + \frac{216}{6} + \frac{1296}{24} + \frac{7776}{120} ight]$$= e^{-6} [6 + 18 + 36 + 54 + 64.8]$$= e^{-6} [178.8]$$= 6 	imes e^{-6} \left( \frac{178.8}{6} ight) = 6 	imes e^{-6} (29.8)$.

$X = x_i$	$3$	$5$	$7$	$9$
$P(X = x_i)$	$k$	$2k$	$3k$	$4k$

The number of persons joining a cinema ticket counter in a minute follows a Poisson distribution with parameter $\lambda = 6$. The probability that at least one and at most five persons join the queue in a particular minute is:

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