Question

The shape of the distribution of the time is 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 4.2 minutes. Complete parts (a) through (c).

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

A. The sample size needs to be less than or equal to 30.

B. The sample size needs to be greater than or equal to 30.

C. The normal model cannot be used if the shape of the distribution is unknown.

D. Any sample size could be used.

(b) What is the probability that a random sample of n=40 oil changes results in a sample mean time less than 15 minutes?

The probability is approximately _. (round to four decimal places as needed.)

(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 40 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

a)

B. The sample size needs to be greater than or equal to 30

b)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	16.2
std deviation   =σ=	4.2000
sample size       =n=	40
std error=σx̅=σ/√n=	0.6641

probability that a random sample of n=40 oil changes results in a sample mean time less than 15 minutes :

probability =

P(X<15)

=

P(Z<-1.81)=

0.0351

c)

for 10th percentile critical value of z=		-1.28
therefore corresponding value=mean+z*std deviation=			15.3 minutes