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jok3333 [9.3K]
3 years ago
12

According to records from a large public university, 88% of students who graduate from the university successfully find employme

nt in their chosen field within three months of graduation. What is the probability that of nine randomly selected students who have graduated from this university, at least six of them find employment in their chosen field within three months
Mathematics
1 answer:
erastovalidia [21]3 years ago
4 0

Answer:

0.9842 = 98.42% probability that at least six of them find employment in their chosen field within three months.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they found employment, or they did not. The probability of a student finding employment is independent of any other student, 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.

88% of students who graduate from the university successfully find employment in their chosen field within three months of graduation.

This means that p = 0.88

Nine randomly selected students

This means that n = 9

What is the probability that of nine randomly selected students who have graduated from this university, at least six of them find employment in their chosen field within three months?

This is:

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)

In which

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

P(X = 6) = C_{9,6}.(0.88)^{6}.(0.12)^{3} = 0.0674

P(X = 7) = C_{9,7}.(0.88)^{7}.(0.12)^{2} = 0.2119

P(X = 8) = C_{9,8}.(0.88)^{8}.(0.12)^{1} = 0.3884

P(X = 9) = C_{9,9}.(0.88)^{9}.(0.12)^{0} = 0.3165

Then

P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) = 0.0674 + 0.2119 + 0.3884 + 0.3165 = 0.9842

0.9842 = 98.42% probability that at least six of them find employment in their chosen field within three months.

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