Question

A study indicates that 62% of students have have a laptop. You randomly sample 8 students. Find the probability that between 4 and 6 (including endpoints) have a laptop.

Answer 1

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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

A study indicates that 62% of students have have a laptop.

This means that $n = 0.62$

You randomly sample 8 students.

This means that $n = 8$

Find the probability that between 4 and 6 (including endpoints) have a laptop.

$P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 4) = C_{8,4}.(0.62)^{4}.(0.38)^{4} = 0.2157$

$P(X = 5) = C_{8,5}.(0.62)^{5}.(0.38)^{3} = 0.2815$

$P(X = 6) = C_{8,6}.(0.62)^{6}.(0.38)^{2} = 0.2297$

$P(4 \leq X \leq 6) = P(X = 4) + P(X = 5) + P(X = 6) = 0.2157 + 0.2815 + 0.2297 = 0.7269$

72.69% probability that between 4 and 6 (including endpoints) have a laptop.