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 16.5 minutes​, and the standard deviation is 4.3 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.

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 = = 16.5 minutes

standard deviation = = 4.3 minutes

n = 35

=    = 16.5 minutes

= / n = 4.3 / 35 = 0.727

Using standard normal table,

P(Z < z) = 10%

= P(Z < z ) = 0.10

= P(Z < -1.28 ) = 0.10  

z = -1.28

Using z-score formula  

= z * +   

= -1.28 * 0.727 + 16.5

= 15.6 minutes

There is 10% chance of being at or below a mean oil-change time of 15.6  minutes.