Question

The time taken by an automobile mechanic to complete an oil change is random with mean 29.5 minutes and standard deviation 3 minutes.

a. What is the probability that 50 oil changes take more than 1500 minutes?

b. What is the probability that a mechanic can complete 80 or more oil changes in 40 hours?

c. The mechanic wants to reduce the mean time per oil change so that the probability is 0.99 that 80 or more oil changes can be completed in 40 hours. What does the mean time need to be? Assume the standard deviation remains 3 minutes.

Answer 1

Let X be a normal random variable which denotes the time required for the mechanic for oli change

Mean, = 29.5 minutes

Standard deviations l, = 3 minutes

(a) For 50 oil changes, the average time required = = 1475 minutes

The standard deviation = = 21.213 minutes

Corresponding to 1500 minutes, the z value = (1500 - 1475)/21.213

= 1.179

Thus, the probability that 50 oil changes take more than 1500 minutes

= P(Z > 1.179) = 0.1193

(b) A mechanic can complete 80 or more oil changes in 40 hours implies that mechanic takes less than 40 hours to complete 80 oil changes

Mean time for 80 oil changes = 29.5*80 = 2360 minutes

Standard deviation = = 26.833 minutes

Corresponding to 40 hours, i.e 2400 minutes, the z value

= (2400 - 2360)/26.833 = 1.49

The required probability = P(Time < 2400)

= P(Z < 1.49) = 0.9319

(c) Corresponding to probability of 0.99, the z value should be 2.327

Thus,

->

Thus, the mean time need to be 29.22 minutes