Question

An instant oil change shop has two bays in which to do oil changes. Each oil change takes 15 minutes to complete. For a 15-minute time period, the owner of the business estimates that the shop is empty 20% of the time and full 30% of the time. Let represent the number of vehicles that are being serviced during a given 15-minute time period.

(a) Write the probability distribution for this situation.

Answer 1

(b) What is the expected number of vehicles being serviced in a given 15-minute time period?

Answer 2

can't open attachment. is this poisson distribution?

Answer 3

I would say the probability distribution is as follows:
0 cars being serviced: 0.2
1 car being serviced: 0.5
2 cars being serviced: .3
No?

Answer 4

Oh. And we couldn't see the name of the variable. It dropped out when you c&p it. But whatever that variable is, that's what the distribution is for.

Answer 5

So, supposing the variable is n (for number of cars) then the probability distribution for n is
0: .2
1: .5
2: .3

and then to calculate expected value, just multiply the probability of each possible n by the outcome in the case of that n and add the results.

0.2(0) + 0.5(1) + 0.3(2)

Answer 6

so 1.1?

Answer 7

what is (B)?

Answer 8

would that be 1.1?

Answer 9

Sorry, yes.

Answer 10

In general, for expected value, if you have n possible outcomes, the expected value is the summation of:
p(n) * f(n)
Where p(n) is the probability of n, and f(n) is the outcome you care about, assuming choice n happens. In this case, you care about the number of cars, so it's just n. If they had asked about the amount of money you made, then f(n) would be the cost of an oil change multiplied by n.