Mathematics: The distribution of customers arriving at a bank is Poisson with a standard deviation of 2 customers per 15-minutes. What is the probability that more than 3 customers arrive during 15 minutes

The answer would be c.hope this ....

The distribution of customers arriving at a bank

kaitlynmoore42

21.01.2022

56.65% probability that more than 3 customers arrive during 15 minutes

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval, which is the same as the variance. The standard deviation is the square root of the variance.

Standard deviation of 2 customers per 15-minutes.

So $\mu = 2^{2} = 4$

What is the probability that more than 3 customers arrive during 15 minutes

Either three or less customers arrive, or more than 3 do. The sum of these probabilities is decimal 1. Mathematically, we have that:

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

We want P(X > 3). So

$P(X 3) = 1 - P(X \leq 3)$

In which

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.0183$

$P(X = 1) = \frac{e^{-4}*(4)^{1}}{(1)!} = 0.0733$

$P(X = 2) = \frac{e^{-4}*(4)^{2}}{(2)!} = 0.1465$

$P(X = 3) = \frac{e^{-4}*(4)^{3}}{(3)!} = 0.1954$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335$

$P(X 3) = 1 - P(X \leq 3) = 1 - 0.4335 = 0.5665$

56.65% probability that more than 3 customers arrive during 15 minutes

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