Question

At a local community college, 57% of students who enter the college as freshmen go on to graduate. Five freshmen are randomly selected.
a. What is the probability that none of them graduates from the local community college? (Do not round intermediate calculations Round your final answer to 4 decimal places Probability
b. What is the probability that at most four will graduate from the local community college? (Do not round intermediate calculations. Round your final answer to 4 decimal places.)
c. What is the expected number that will graduate? (Round your final answer to 2 decimal places)

Answer 1

Answer:

a) 0.0147 = 1.47% probability that none of them graduates from the local community college.

b) 0.9398 = 93.98% probability that at most four will graduate from the local community college.

c) The expected number that will graduate is 2.85.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they will graduate, or they will not. The probability of a student graduating is independent of any other student graduating, which means that the binomial probability distribution is used 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.

57% of students who enter the college as freshmen go on to graduate.

This means that $p = 0.57$

Five freshmen are randomly selected.

This means that $n = 5$

a. What is the probability that none of them graduates from the local community college?

This is P(X = 0). So

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

$P(X = 0) = C_{5,0}.(0.57)^{0}.(0.43)^{5} = 0.0147$

0.0147 = 1.47% probability that none of them graduates from the local community college.

b. What is the probability that at most four will graduate from the local community college?

This is:

$P(X \leq 4) = 1 - P(X = 5)$

In which

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

$P(X = 5) = C_{5,5}.(0.57)^{5}.(0.43)^{0} = 0.0602$

So

$P(X \leq 4) = 1 - P(X = 5) = 1 - 0.0602 = 0.9398$

0.9398 = 93.98% probability that at most four will graduate from the local community college.

c. What is the expected number that will graduate?

The expected value of the binomial distribution is:

$E(X) = np$

In this question:

$E(X) = 5*0.57 = 2.85$

The expected number that will graduate is 2.85.