Answer:
0.5940 = 59.40% probability that at least two customers are there for the next hour.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval.
On average, the shoppers across McMaster Univerisity have 2 customers per hour
This means that [tex]\mu = 2[/tex]
Find the probability that at least two customers are there for the next hour.
This is:
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]. So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353[/tex]
[tex]P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707[/tex]
[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.1353 + 0.2707 = 0.4060[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.4060 = 0.5940[/tex]
0.5940 = 59.40% probability that at least two customers are there for the next hour.