Question

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 21.6 minutes and the standard deviation is 4.7 minutes.

Complete parts ((c).

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

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

Answer #1

Even though the shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is unknown. since sample size = n = 45 > 30 Large Sample and Population Standard Deviation = = 4.7 is provided, by Central Limit Theorem, the Sampling Distribution of Sample means is Normal Distribution.

= 21.6

= 4.7

n = 45

SE = /

= 4.7/

= 0.7006

Minimum 10% corresponds to area = 0.50 - 0.10 = 0.40 frommid value to Z on LHS.

Table of Area Under Standard Normal Curve gives Z = - 1.28

So,

we get

Z = - 1.28 = (
- 21.6)/0.7006

So,

= 21.6 - (1.28 X 0.7006)

= 20.7032

So,

Answer is:

**20.7032**

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