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