Question

The shape of the distribution of the time required to get an oil change at a

1515 -minute

oil-change facility is unknown. However, records indicate that the mean time is

16.6 minutes

and the standard deviation is

3.6 minutes

Complete parts (a) through (c) below.

What is the probability that a random sample of

nequals=35

oil changes results in a sample mean time less than

15

minutes?

Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform

35

oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the manager.

Answer 1

By Central limit theorem, the sample mean with larger sample size (n > 30) will follow normal distribution.

Mean = Sample mean = 16.6 minutes

Standard deviation of sample mean = 3.6 / = 0.6085

Thus,   ~ Normal(16.6, 0.6085²)

Probability that sample mean time less than 15 minutes = P( < 15)

= P[Z < (16.6 - 15) / 0.6085]

= P[Z < 2.63]

= 0.9957

Z value for p = 10% = -1.28

Mean oil-change time at 10% chance of being at or below =

= 16.6 - 1.28 * 0.6085

= 15.82 minutes