A public interest group hires students to solicit donations by telephone. After

Question

A public interest group hires students to solicit donations by telephone. After a brief training period students make calls to potential donors and are paid on a commission basis. Experience indicates that early on, these students tend to have only modest success and that 90 % of them give up their jobs in their first two weeks of employment. The group hires 5 students, which can be viewed as a random sample. What is the probability that at least 2 of the 5 will not give up in the first two weeks?

Explanation / Answer

this follows binomial distribution

p = proportion of students who dont give up = 1 -90% = 0.1

P(X>=2)

= 1 - P(X=0) - P(X=1)

= 1 - 5C0 * 0.9^5 - 5C1 * 0.1 * 0.9^4

= 1 - 0.59049 - 0.32805

= 0.08146

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