Respuesta :
Answer:
99.1% probability that at most 8 of the calls involve a fax message
Step-by-step explanation:
For each call, there are only two possible outcomes. Either it is a fax message, or it is not. The probability of a call being a fax message is independent from other calls. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Suppose that 20% of the incoming calls involve fax messages, and consider a sample of 20 incoming calls
This means that [tex]p = 0.2, n = 20[/tex]
(a) What is the probability that at most 8 of the calls involve a fax message
[tex]P(X \leq 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.012[/tex]
[tex]P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.058[/tex]
[tex]P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.137[/tex]
[tex]P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.205[/tex]
[tex]P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.218[/tex]
[tex]P(X = 5) = C_{20,0}.(0.2)^{5}.(0.8)^{15} = 0.175[/tex]
[tex]P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.109[/tex]
[tex]P(X = 7) = C_{20,7}.(0.2)^{7}.(0.8)^{13} = 0.055[/tex]
[tex]P(X = 8) = C_{20,8}.(0.2)^{8}.(0.8)^{12} = 0.022[/tex]
[tex]P(X \leq 8) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) = 0.012 + 0.058 + 0.137 + 0.205 + 0.218 + 0.175 + 0.109 + 0.055 + 0.022 = 0.991[/tex]
99.1% probability that at most 8 of the calls involve a fax message
