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Sagot :
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
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
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 [tex]n = 10[/tex]
- Suppose that 0.3 of them have to wait more than 8 minutes, hence [tex]p = 0.3[/tex]
The probability that at most 2 of the next 10 callers will have to wait more than 8 minutes is:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
Then
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.3)^{0}.(0.7)^{10} = 0.0282[/tex]
[tex]P(X = 1) = C_{10,1}.(0.3)^{1}.(0.7)^{9} = 0.1211[/tex]
[tex]P(X = 2) = C_{10,2}.(0.3)^{2}.(0.7)^{8} = 0.2335[/tex]
Then:
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0282 + 0.1211 + 0.2335 = 0.3828[/tex]
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 https://brainly.com/question/25537909
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