Question

Records show that the average number of phone calls received per day is 9.2. Find the probability of between 2 and 4 phone calls received (endpoints included) in a given day.

Answer 1

Answer:

4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.

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.

Records show that the average number of phone calls received per day is 9.2.

This means that $\mu = 9.2$.

Find the probability of between 2 and 4 phone calls received (endpoints included) in a given day.

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

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

$P(X = 2) = \frac{e^{-9.2}*(9.2)^{2}}{(2)!} = 0.0043$

$P(X = 3) = \frac{e^{-9.2}*(9.2)^{3}}{(3)!} = 0.0131$

$P(X = 4) = \frac{e^{-9.2}*(9.2)^{4}}{(4)!} = 0.0302$

$(2 \leq X \leq 4) = P(X = 2) + P(X = 3) + P(X = 4) = 0.0043 + 0.0131 + 0.0302 = 0.0476$

4.76% probability of between 2 and 4 phone calls received (endpoints included) in a given day.