Question

suppose a fast food restaurant can expect two customers every three minutes on the average.what is the probability that four or fewer patrons will enter the restaurant 9minute period?On a given day​

Answer 1

Answer:

The required answer is $0.2851$.

Step-by-step explanation:

Poisson Distribution Formula:

$P(X=x \mid \lambda)=\frac{e^{-\lambda} \lambda^{x}}{x !}$

$X \sim \text { Poisson }(\lambda)$

1. Mean, $\mu=\lambda$

2. Variance, $\quad \sigma^{2}=\lambda$

3. Standard deviation, $\sigma=\sqrt{\lambda}$

where:

$x=$ quantity of activities in a potential area.

$\lambda=$ expected number of events

$e=$ base of the natural logarithm system $(2.71828 \ldots)$.

The probability of exactly ' $x$' occurrences is equal to:

$P(X=x \mid \lambda)=\frac{e^{-\lambda} \lambda^{x}}{x !}$

Given: $\lambda_{3 \text { mins }}=2$ customers

$\lambda_{9 \text { mins }}=2 \times 3=6 \text { customers }$

Let $X=$ number of customers enter the restaurant

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

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

$&=0.0025+0.0149+0.0446+0.0892+0.1339$

$&=0.2851$

Therefore, The probability that four or fewer patrons will enter the restaurant in a $9$ minute period is $0.2851$.

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