Question

Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. During a given hour, what are the probabilities that:A: no more than three customers arrive?B: at least two customers arrive?C: exactly five customers arrive?

Answer 1

Answer:

a) $P(X\leq 3) =0.0009119+0.00638+0.0223+0.0521=0.0818$

b) $P(X \geq 2) = 1-P(X$

c) $P(X=5) =\frac{e^{-7} 7^5}{5!}=0.128$

Step-by-step explanation:

Let X the random variable that represent the number of customers that arrive in a given hour. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this case we have $\lambda= 7 \frac{customers}{hour}$

Part a

No more than 3 means:

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

And if we find the individual probabilites we got:

$P(X=0) =\frac{e^{-7} 7^0}{0!}=0.0009119$

$P(X=1) =\frac{e^{-7} 7^1}{1!}=0.00638$

$P(X=2) =\frac{e^{-7} 7^2}{2!}=0.0223$

$P(X=3) =\frac{e^{-7} 7^3}{3!}=0.0521$

So then if we replace we got:

$P(X\leq 3) =0.0009119+0.00638+0.0223+0.0521=0.0818$

Part b

We want the probability that at least two so we can do this:

$P(X \geq 2) = 1-P(X$

The last step by the complement rule.

$P(X=0) =\frac{e^{-7} 7^0}{0!}=0.0009119$

$P(X=1) =\frac{e^{-7} 7^1}{1!}=0.00638$

$P(X \geq 2) = 1-P(X$

Part c

$P(X=5) =\frac{e^{-7} 7^5}{5!}=0.128$