Question

The shape of the distribution of the time required to get an oil change at a 10 -minute oil-change facility is unknown. However, records indicate that the meantime is 11.8 minutes , and the standard deviation is 4.6 minutes.Complete parts (a) through (c) below.

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

Choose the required sample size below.

A.

The sample size needs to be greater than 30.

B.

Any sample size could be used.

C.

The sample size needs to be less than 30.

D.

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

(b) What is the probability that a random sample of

nequals=4545

oil changes results in a sample mean time less than

1010

minutes?The probability is approximately

nothing.

(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

4545

oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below? This will be the goal established by the manager.There would be a 10% chance of being at or below

nothing

minutes.

(Round to one decimal place as needed.)

Answer 1

a)

A.

The sample size needs to be greater than 30.

b)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	11.8
std deviation   =σ=	4.6000
sample size       =n=	45
std error=σx̅=σ/√n=	0.6857

probability =

P(X<10)

=

P(Z<-2.62)=

0.0044

c_)

for 10th percentile critical value of z=		-1.28
therefore corresponding value=mean+z*std deviation=			10.9 minutes