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.

There are 45 oil changes between 10 a.m. and 12 p.m. treating this as a random sample, n = 45 and there would be a 10% chance of the mean oil-change time, therefore P = 10% = 0.1

A probability of 0.1 gives a corresponding z score of -1.28, this is gotten from the z table. z = -1.28

z score (z) = (x - mean) / (standard deviation / √n)

Let us assume mean = 21.2 minutes and standard deviation = 3.5 minutes. substituting values:

-1.28 = (x - 21.2)/(3.5 ÷ √45)

-1.28 = (x - 21.2) / 0.522

x - 21.2 = -0.6678

x = 21.2 - 0.6678 = 20.53

x = 20.53

andromache andromache · Answer 2 · 2022-04-05T14:59:04+02:00

There is a 10% chance of the mean oil-change time of 20.53 minutes.

Calculation of the time taken:

Since

There are 45 oil changes between 10 a.m. and 12 p.m. treating this as a random sample, n = 45 and there would be a 10% chance of the mean oil-change time, therefore P = 10% = 0.1

Now the z score should be -1.28

Now

z score (z) = (x - mean) / (standard deviation / √n)

here we assume mean = 21.2 minutes and standard deviation = 3.5 minutes

So,

-1.28 = (x - 21.2)/(3.5 ÷ √45)

-1.28 = (x - 21.2) / 0.522

x - 21.2 = -0.6678

x = 21.2 - 0.6678 = 20.53

x = 20.53

hence, There is a 10% chance of the mean oil-change time of 20.53 minutes.

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