Question

based on the simulation, what is the probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered?

Answer 1

Using the binomial distribution, supposing that 0.3 of the callers have to wait more than 8 minutes to have their calls answered, it is found that there is a 0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

For each caller, there are only two possible outcomes, either they have to wait more than 8 minutes to have their calls answered, or they do not. The probability of a caller having to wait more than 8 minutes is independent of any other caller, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• n is the number of trials.
• p is the probability of a success on a single trial.

In this problem:

• 10 callers, hence $n = 10$
• Suppose that 0.3 of them have to wait more than 8 minutes, hence $p = 0.3$

The probability that at most 2 of the next 10 callers will have to wait more than 8 minutes is:

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

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.3)^{0}.(0.7)^{10} = 0.0282$

$P(X = 1) = C_{10,1}.(0.3)^{1}.(0.7)^{9} = 0.1211$

$P(X = 2) = C_{10,2}.(0.3)^{2}.(0.7)^{8} = 0.2335$

Then:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0282 + 0.1211 + 0.2335 = 0.3828$

0.3828 = 38.28% probability that at most 2 of the next 10 callers will have to wait more than 8 minutes to have their calls answered.

A similar problem is given at brainly.com/question/25537909