A large stockpile of used pumps contains 20% that are currently unusable and need to be repaired. A repairman is sent to the stockpile with three repair kits. He selects pumps at random and tests them one at a time. If a pump works, he goes on to the next one. If a pump doesn't work, he uses one of his repair kits on it. Suppose that it takes 10 minutes to test whether a pump works, and 20 minutes to repair a pump that does not work. The expected value and variance of the total time it takes the repairman to use up his three kits are, respectively:.

Expected value = 190

Variance = 4000

Step-by-step explanation:

Let X be the number of the trials until the third success of the bad pump.

This implies that X is a negative binomial distribution having θ = 20% = 0.2.

Now, if for example it will take X trials to use up the three pumps, then the total time is 10 min/trials + extra 10 minutes for the 3 bad pumps

This means the total time is written as;

T = 10X + (10 + 10 + 20)

T = 10X + 40

Mean which is also expected value of X is;

μ_x = 3/0.2 = 15

Variance of X is; σ²_x = 40

Thus;

Mean of T will be;

μ_T = 10μ_x + 40

μ_T = 10(15) + 40

μ_T = 190

Also, variance of T will be;

σ²_T = 10²•σ²_x

σ²_T = 100 × 40

σ²_T = 4000

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