Question

The shape of the distribution of the time required to get an oil change at a

1010​-minute

​oil-change facility is unknown.​ However, records indicate that the mean time is

11.6 minutes11.6 minutes​,

and the standard deviation is

4.5 minutes4.5 minutes.

Complete parts​ (a) through​ (c) below.

Click here to view the standard normal distribution table (page 1).

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Click here to view the standard normal distribution table (page 2).

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​(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 less than 30.

B.

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

C.

The sample size needs to be greater than 30.

D.

Any sample size could be used.

​(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

​(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.

Answer 1

Solution :

Given that ,

mean = = 11.6 minutes

standard deviation = = 4.5 minutes

n = 45

a) correct option is = C

The sample size needs to be greater than 30.

=    = 11.6 minutes

= / n = 4.5 / 45 = 0.671

b) P( < 10) = P(( - ) / < (10 - 11.6) / 0.671)

= P(z < -2.38)

Using z table

= 0.0087

c) 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 * 0.671 + 11.6

= 10.7 minutes

There is 10% chance of being at or below a mean oil-change time of 10.7 minutes