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 21.2 minutes​, and the standard deviation is 3.4 minutes.

Complete parts ​(a) through (c).

​(a) To compute probabilities regarding the sample mean using the normal​ model, what size sample would be​ required?

​(b) What is the probability that a random sample of n=45 oil changes results in a sample mean time less than 20 ​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 45 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

Solution :

Given that ,

mean = = 21.2

standard deviation = = 3.4

n = 45

a) The sample size needs to be greater than or equal 30

=   = 21.2

= / n = 3.4 / 45 = 0.507

b) P( < 20) = P(( - ) / < (20 - 21.2) / 0.507)

= P(z < -2.37)

Using z table

= 0.0089

c) 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 * 0.507 + 21.2

= 20.6 minutes