# Getting the mean without having the total population

I have the following question:

$$10$$% of applications for a job possess the right skills. A company has 3 positions to fill and they interview applicants one at a time until they fill all $$3$$ positions. The company takes $$3$$ hours to interview an unqualified applicant and $$5$$ hours to interview a qualified applicant. Calculate the mean of the time to conduct all interviews.

I got a bit lost considering that they wanted to mean of all interviews however I just did $$0.10*3+0.90*5=3.2$$ which seemed too simplistic, any help in correction/verification would be appreciated.

• Maybe a good idea to start with a (slightly) simpler problem. Suppose the firm only needs to hire one (qualified) applicant. How many hours is that expected to take?
– lulu
Nov 30 '20 at 15:05

First, we can note that, no matter how many applicants are interviewed, $$3$$ qualified applicants will be interviewed, taking a total of $$3*5=15$$ hours.

Next, we need to find how many unqualified applicants are interviewed. Since each applicant can be assumed to be independently equally likely to be qualified, and you stop interviewing after a fixed number of qualified applicants, the number of unqualified applicants interviewed is given by a Negative Binomial random variable, with parameters $$p=0.9$$ (the probability of an unqualified candidate) and $$r=3$$ (the number of qualified candidates). The mean of this is given by $$\dfrac{pr}{1-p}=\dfrac{0.9*3}{1-0.9}=27$$.

Therefore, the mean time taken for all interviews is $$3$$ hours for each unqualified candidate plus $$5$$ hours for each qualified candidate, giving a total of $$27*3 + 3*5 = 96$$

• thank you, alll clear! Nov 30 '20 at 15:08

I think you were assuming that only one interview was done.

What happens if the company conducts interviews until 3 qualified applicants are found ?

• No idea how to do this since the total number isn't mentioned, maybe 0.10*3*5+0.9*3? Nov 30 '20 at 14:56