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.

Incomplete question

First aspect of the question is typed below.

The shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility

is unknown. However, records indicate that the mean time is 2l.2-minutes, and the standard deviation 3.5 minutes.

Explanation:

Employees bonus $50

35 oil changes between 10:00 am to 12:00pm

n = 35

10% changes

So, the z - score can be calculated using

z-score = InvNorm(0.10)

z-score = -1.28

So, given that,

Standard deviation is 3.5minutes

σ = 3.5 minutes

Mean time is 21.2 minutes

μx = 21.2 minutes

Then,

σx = σ / √n

σx = 3.5 / √35

σx = 0.5916 minutes

Then, Z score can be written as

Z = (x - μx) / σx

-1.28 =  (x - 21.2) / 0.5916

Cross multiply

-1.28 × 0.5916 = x - 21.2

-0.7573 = x - 21.2

x = 21.2 - 0.7573

x = 20.443 minutes

There is a 10% chance of being at or below a mean oil-change time of 20.44 minutes

answer; tech (a) is correct;

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