Question

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 4545 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. There would be a​ 10% chance of being at or below nothing minutes. ​(Round to one decimal place as​ needed.)

Answer 1

Answer:

The answer is given below

Step-by-step explanation:

The mean (μ) = 21.2 minutes and the standard deviation σ = 3.5 minutes.

the​ oil-change facility will perform 45 oil changes between 10 A.M. and 12 P.M, therefore the sample size n = 45

there be a​ 10% chance of being at or​ below. From the normal distribution table, The z score corresponding to a probability of 10% (= 0.1)  is -1.28.

z = -1.28

To calculate the mean​ oil-change time, we use the formula:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n} } }\\\\ Substituting \ values:\\-1.28=\frac{x-21.2}{\frac{3.5}{\sqrt{45} } }\\\\x-21.2=-0.6678\\\\x=-0/6678+21.2\\\\x=20.5$