Respuesta :
Answer:
a) 0.968 = 96.8% probability that at most 7 of the calls involve a fax message
b) 0.055 = 5.5% probability that exactly 7 of the calls involve a fax message
c) 0.913 = 91.3% probability that at least 7 of the calls involve a fax message
d) 0.032 = 3.2% probability that more than 7 of the calls involve a fax message
Step-by-step explanation:
For each call, there are only two possible outcomes. Either there is a fax message, or there is not. The probability of a call being a fax 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.
In this problem we have that:
[tex]p = 0.2, n = 20[/tex]
(a) What is the probability that at most 7 of the calls involve a fax message?
This is [tex]P(X \leq 7)[/tex].
So
[tex]P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{20,0}.(0.2)^{0}.(0.80)^{20} = 0.0115[/tex]
[tex]P(X = 1) = C_{20,1}.(0.2)^{1}.(0.80)^{19} = 0.0576[/tex]
[tex]P(X = 2) = C_{20,2}.(0.2)^{2}.(0.80)^{18} = 0.1369[/tex]
[tex]P(X = 3) = C_{20,3}.(0.2)^{3}.(0.80)^{17} = 0.2054[/tex]
[tex]P(X = 4) = C_{20,4}.(0.2)^{4}.(0.80)^{16} = 0.2182[/tex]
[tex]P(X = 5) = C_{20,5}.(0.2)^{5}.(0.80)^{15} = 0.1746[/tex]
[tex]P(X = 6) = C_{20,6}.(0.2)^{6}.(0.80)^{14} = 0.1091[/tex]
[tex]P(X = 7) = C_{20,7}.(0.2)^{7}.(0.80)^{13} = 0.0545[/tex]
[tex]P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 + 0.0545 = 0.968[/tex]
0.968 = 96.8% probability that at most 7 of the calls involve a fax message
(b) What is the probability that exactly 7 of the calls involve a fax message?
[tex]P(X = 7) = C_{20,7}.(0.2)^{7}.(0.80)^{13} = 0.0545[/tex]
Rounded to three decimal places
0.055 = 5.5% probability that exactly 7 of the calls involve a fax message
(c) What is the probability that at least 7 of the calls involve a fax message?
Either less than 7 of the calls involve a fax message, or at least 7 does. The sum of the probabilities of these events is 1. So
[tex]P(X < 7) + P(X \geq 7) = 1[/tex]
We want [tex]P(X \geq 7)[/tex]. So
[tex]P(X \geq 7) = 1 - P(X < 7)[/tex]
In which
[tex]P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.913[/tex]
0.913 = 91.3% probability that at least 7 of the calls involve a fax message
(d) What is the probability that more than 7 of the calls involve a fax message?
Either 7 or less calls involve a fax message, or at least 7 do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 7) + P(X > 7) = 1[/tex]
We want P(X > 7).
From a, we have that [tex]P(X \leq 7) = 0.968[/tex]. So
[tex]P(X \leq 7) + P(X > 7) = 1[/tex]
[tex]0.968 + P(X > 7) = 1[/tex]
[tex]P(X > 7) = 0.032[/tex]
0.032 = 3.2% probability that more than 7 of the calls involve a fax message
