Respuesta :
Answer:
0.5628 = 56.28% probability that at least eight of the 40 subjects revert to their first learned method.
Step-by-step explanation:
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]n = 40, p = 0.2[/tex]
Find the probability that at least eight of the 40 subjects revert to their first learned method.
Either less than 8 do, or at least 8 do. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 8) + P(X \geq 8) = 1[/tex]
We want [tex]P(X \geq 8)[/tex]. So
[tex]P(X \geq 8) = 1 - P(X < 8)[/tex]
In which
[tex]P(X < 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)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{40,0}.(0.2)^{0}.(0.8)^{40} = 0.0001[/tex]
[tex]P(X = 1) = C_{40,1}.(0.2)^{1}.(0.8)^{39} = 0.0013[/tex]
[tex]P(X = 2) = C_{40,2}.(0.2)^{2}.(0.8)^{38} = 0.0065[/tex]
[tex]P(X = 3) = C_{40,3}.(0.2)^{3}.(0.8)^{37} = 0.0205[/tex]
[tex]P(X = 4) = C_{40,4}.(0.2)^{4}.(0.8)^{36} = 0.0475[/tex]
[tex]P(X = 5) = C_{40,5}.(0.2)^{5}.(0.8)^{35} = 0.0854[/tex]
[tex]P(X = 6) = C_{40,6}.(0.2)^{6}.(0.8)^{34} = 0.1246[/tex]
[tex]P(X = 7) = C_{40,7}.(0.2)^{7}.(0.8)^{33} = 0.1513[/tex]
[tex]P(X < 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) = 0.0001 + 0.0013 + 0.0065 + 0.0205 + 0.0475 + 0.0854 + 0.1246 + 0.1513 = 0.4372[/tex]
[tex]P(X \geq 8) = 1 - P(X < 8) = 1 - 0.4372 = 0.5628[/tex]
0.5628 = 56.28% probability that at least eight of the 40 subjects revert to their first learned method.
