Respuesta :
Answer:
72.69% probability that between 4 and 6 (including endpoints) have a laptop.
Step-by-step explanation:
For each student, there are only two possible outcomes. Either they have a laptop, or they do not. The probability of a student having a laptop is independent from other students. 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.
A study indicates that 62% of students have have a laptop.
This means that [tex]n = 0.62[/tex]
You randomly sample 8 students.
This means that [tex]n = 8[/tex]
Find the probability that between 4 and 6 (including endpoints) have a laptop.
[tex]P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{8,4}.(0.62)^{4}.(0.38)^{4} = 0.2157[/tex]
[tex]P(X = 5) = C_{8,5}.(0.62)^{5}.(0.38)^{3} = 0.2815[/tex]
[tex]P(X = 6) = C_{8,6}.(0.62)^{6}.(0.38)^{2} = 0.2297[/tex]
[tex]P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.2157 + 0.2815 + 0.2297 = 0.7269[/tex]
72.69% probability that between 4 and 6 (including endpoints) have a laptop.
