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

Explanation:

Given that:

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

answer; i believe the correct answer is (180)

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