Question

The shape of the distribution of the time required to get an oil change at a 15-minute oil-change facility is unknown. However, records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.
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 35 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

Answer:

The mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.

Step-by-step explanation:

We are given that records indicate that the mean time is 16.2 minutes, and the standard deviation is 3.4 minutes.

On a typical​ Saturday, the​ oil-change facility will perform 35 oil changes between 10 A.M. and 12 P.M. Assuming that data follows normal distribution.

Let $\bar X$ = sample mean time

The z score probability distribution for sample mean is given by;

                          Z  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\mu$ = population mean time = 16.2 minutes

            $\sigma$ = standard deviation = 3.4 minutes

            n = sample size = 35 oil changes

Now, the mean oil-change time that would there be a 10% chance of being at or below is given by;

          P(X $\leq$ x) = 0.10          {where x is required mean oil-change time}

          P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }$ ) = 0.10

          P(Z $\leq$ $\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }$ ) = 0.10

Now, in the z table the critical value of X which represents the below 10% of the probability area is given as -1.282, that means;

                  $\frac{x-16.2}{\frac{3.4}{\sqrt{35} } }$  =  -1.282

               x - 16.2  =  $-1.282 \times {\frac{3.4}{\sqrt{35} } }$

                         x  =  16.2 - 0.74 = 15.46 minutes

Hence, the mean oil-change time that would there be a 10% chance of being at or below is 15.46 minutes.