Question

The shape of the distribution of the time required to get an oil change at a 20​-minute ​oil-change facility is unknown.​ However, records indicate that the mean time is 21.4 minutes​, and the standard deviation is 4.5 minutes.

​(c) 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.

There is a​ 10% chance of being at or below a mean​ oil-change time of ___ minutes.

​(Round to one decimal place as​ needed.)

Answer 1

Solution,

Given that,

mean = = 21.4

standard deviation = = 4.5

n = 35

= = 21.4

= / n = 4.5 / 35 = 0.76

Using standard normal table

P(Z < z ) = 10%

P(Z < z ) = 0.10

P(Z < -1.282 ) = 0.10

z = -1.282

Using z-score formula

= z * +

= -1.282 * 0.76 + 21.4

= 20.4 minutes.