The number of telephone calls that arrive at a phone exchange is often modeled as a Poisson random variable. Assume that on the average there are 10 calls per hour. (a) What is the probability that there are three or fewer calls in one hour

the probability that there are three or fewer calls in one hour is 0.011

Step-by-step explanation:

The formula for poisson distribution is expressed as

P(x = r) = (e^- µ × µ^r)/r!

Where

µ represents the mean of the theoretical distribution.

r represents the number of successes of the event.

From the information given,

µ = 10

For the probability that there are three or fewer calls in one hour, it is expressed as

P(x ≤ 3) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3)

Therefore,

P(x = 0) = (e^- 10 × 10^0)/0! = 0.000045

P(x = 1) = (e^- 10 × 10^1)/1! = 0.00045

P(x = 2) = (e^- 10 × 10^2)/2! = 0.0023

P(x = 3) = (e^- 10 × 10^3)/3! = 0.0077

Therefore,

P(x ≤ 3) = 0.000045 + 0.00045 + 0.0023 + 0.0077 = 0.011

i have no idea just need points sorry

step-by-step explanation: