Suppose a fast-food restaurant can expect two customers every 3 minutes, on average. What is the probability that four or fewer customers will enter the restaurant in a 9-minute period?

Given that, a food restaurant can expect two custiomers every 3 minutes. That means, 6 customers in every 9 minutes.Let the random variable x br the number of patroons enter the restaurant in a 9 minute period and x follows poisson ditribution with rate 6.Then the probability mass function of x is,P(X=x)=e-6&nbsp;6xx!,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;x=0,1,2,...∴&nbsp;The probability that four or fewer patroons will enter the restaurant in a 9 minute period is,P(X≤4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=e-6&nbsp;600!+e-6&nbsp;611!+e-6&nbsp;622!+e-6&nbsp;633!+e-6&nbsp;644!=e-6&nbsp;(1+61!+622!+633!+64!)=e-6×115=0.2850565003≈0.2851

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2022-03-28

Suppose a fast-food restaurant can expect two customers every 3 minutes, on average. What is the probability that four or fewer customers will enter the restaurant in a 9-minute period?

Answer & Explanation

Ian Adams

Skilled2022-03-28Added 163 answers

Given that, a food restaurant can expect two custiomers every 3 minutes. That means, 6 customers in every 9 minutes.

Let the random variable x br the number of patroons enter the restaurant in a 9 minute period and x follows poisson ditribution with rate 6.

Then the probability mass function of x is,

$P (X = x) = \frac{e^{- 6} 6^{x}}{x!}, x = 0, 1, 2, . . .$

$∴$ The probability that four or fewer patroons will enter the restaurant in a 9 minute period is,

$P (X \leq 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)$

$= \frac{e^{- 6} 6^{0}}{0!} + \frac{e^{- 6} 6^{1}}{1!} + \frac{e^{- 6} 6^{2}}{2!} + \frac{e^{- 6} 6^{3}}{3!} + \frac{e^{- 6} 6^{4}}{4!}$

$= e^{- 6} (1 + \frac{6}{1!} + \frac{6^{2}}{2!} + \frac{6^{3}}{3!} + \frac{6}{4!})$

$= e^{- 6} \times 115 = 0.2850565003$

$\approx 0.2851$

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