Expected value and Time function

You are conducting interviews.

If the candidate is qualified it takes 30 minutes to interview the candidate.

If the candidate is not qualified it takes 10 minutes to interview the candidate.

a) Create a function for this scenario,

b) then calculate the expected value. The mean is 17.

My best guess on the formula is T = 30y + 10z.

I know this does not work with calculating the expected value because the z needs to be y to do an expected value calculation. Can I please have some ideas with this?

• hint: can't you substitude z with (1-y)? you can't be both qualified and not qualified at the same time after all ;)
– SAJW
Nov 2, 2016 at 2:10
• Can you explain why you plug in 1-y? What happens if you interview 5 unqualified people and 2 qualified people? Nov 2, 2016 at 2:51
• being qualified or not is a probability. since you cant be both at the same time you are to y% qualified and to (100-y)% not qualified. Or to simply remove the % sign, just say 0<y<1 and z=1-y
– SAJW
Nov 2, 2016 at 2:53
• And you plug the 1- y into the z because it's smaller right? Nov 2, 2016 at 3:02
• actually it doesnt matter which variable you substitude with which. The point is: y=1-z AND z=1-y (simple equation solving). The only thing that's for sure is, that 0<y<1, because otherwise the mean would be 10 or 30
– SAJW
Nov 2, 2016 at 3:05

Let $p$ be the probability that the candidate is qualified. Then, the expected time for an interview would be

$$\mathbb E[T] = p 30 + (1-p)10.$$

Therefore, if $\mathbb E[T] = 17$, it must be that

$$p 30 + (1-p)10 = 17$$

which implies that

$$p = 7/20.$$

• Can you explain why you plug in 1-p? What happens if you interview 5 unqualified people and 2 qualified people? Nov 2, 2016 at 2:51
• @cokedude Then $p = 2/7$.
– mzp
Nov 2, 2016 at 3:16
• @mzp p doesnt change that significant in 7 tries. (assuming an almost endless supply of interviewees)
– SAJW
Nov 2, 2016 at 12:26