Suppose we were told by the superintendent of schools that the probability a randomly selected middle school student would have an IQ between 100 and 120 is P=0.68. If we randomly select five students from the eighth grade, what is the probability that three of them will have an IQ between 100 and 120?

Respuesta :

Answer:

32.20% probability that three of them will have an IQ between 100 and 120

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they have an IQ between 100 and 120, or they do not. The students are chosen at random, which means that the probability of a student having IQ between these two values is independent from other students. So we use the binomial probability distribution to solve this problem.

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 = 0.68[/tex]

If we randomly select five students from the eighth grade, what is the probability that three of them will have an IQ between 100 and 120?

This is [tex]P(X = 3)[/tex] when [tex]n = 5[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{5,3}.(0.68)^{3}.(0.32)^{2} = 0.3220[/tex]

32.20% probability that three of them will have an IQ between 100 and 120