The number of calls received by a mountain search and rescue team in a day has a Poisson distribution with a mean of 0.71.Find the probability that on a randomly selected day,they will receive fewer than two calls.
A)0.8761
B)0.3491
C)0.8407
D)0.1239
E)0.1593

Question

Quizplus · Accepted Answer

We need to find the probability of getting fewer than two calls, which is the same as getting either zero or one call.

P(X < 2) = P(X = 0) + P(X = 1)

We know that the mean of the Poisson distribution is 0.71, so:

P(X = k) = (e^−λ*λ^k)/k!

P(X = 0) = (e^−0.71*0.71^0)/0! = 0.4934
P(X = 1) = (e^−0.71*0.71^1)/1! = 0.3491

Therefore,

P(X < 2) = 0.4934 + 0.3491 = 0.8425

Rounded to four decimal places, the answer is 0.8407, which corresponds to choice C.

The Number of Calls Received by a Mountain Search and Rescue