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

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

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 =
`(\bar X-\mu)/((\sigma)/(√(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(
`(\bar X-\mu)/((\sigma)/(√(n) ) )`
`\leq`
`(x-16.2)/((3.4)/(√(35) ) )` ) = 0.10

P(Z
`\leq`
`(x-16.2)/((3.4)/(√(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;

`(x-16.2)/((3.4)/(√(35) ) )` = -1.282

x - 16.2 =
`-1.282 * {(3.4)/(√(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.