Question

A local business promotes oil changes that last on the average of 45 minutes with a standard deviation of 10 minutes.  If the process of changing oil follows a normal distribution, answer the following:

1. What is the probability an oil change takes less than 40 minutes?
2. What is the probability that the oil change takes between 35 and 55 minutes?
3. What is the probability that an oil change takes less than 49 minutes.
4. What is the probability that an oil change takes between 42 and 52 minutes.
5. The company wants to keep its customers.  So for the longest 5% of oil changes, the owner wants to give the customer 50% off his next visit.  How long would it take for the oil change if the customer qualifies for the discount.

Answer 1

Mean, = 45 minutes

Standard deviation, = 10 minutes

P(X < A) = P(Z < (A - )/)

P(an oil change takes less than 40 minutes) = P(X < 40)

= P(Z < (40 - 45)/10)

= P(Z < -0.5)

= 0.3085

P(an oil change takes between 35 and 55 minutes) = P(35 < X < 55)

P(X < 55) - P(X < 35)

= P(Z < (55 - 45)/10) - P(Z < (35 - 45)/10)

= P(Z < 1) + P(Z < -1)

= 0.8413 - 0.1587

= 0.6826

P(an oil change takes less than 49 minutes) = P(X < 49)

= P(Z < (49 - 45)/10)

= P(Z < 0.4)

= 0.6554

P(an oil change takes between 42 and 52 minutes) = P(X < 52) - P(X < 42)

= P(Z < (52 - 45)/10) - P(X < (42 - 45)/10)

= P(Z < 0.7) - P(Z < -0.3)

= 0.7580 - 0.3821

= 0.3759

Let us assume that it would take M minutes for the oil change if the customer qualifies for the discount

P(X > M) = 0.05

P(X < M) = 1 - 0.05 = 0.95

P(Z < (M - 45)/10) = 0.95

Take Z value corresponding to 0.95 from standard normal distribution table

(M - 45)/10 = 1.645

M = 61.45 minutes