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BES Tutorial Sample Solutions, S2/11

TUTORIAL 7

WEEK 8 TUTORIAL EXERCISES (To be discussed in the week starting

September 12)

1. Suppose a normally distributed random variable X has a mean of 50 and a

variance of 100. Also suppose a sample of size 16 is drawn from this

population. Calculate the following probabilities:(a) P( X > 55)

3085.01915.05.0)5.0(

10

5055)55(

=−=>=

−>=>

Z P

Z P X P

(b) ( � > 55)

X ~ )16100,50( N

0228.04772.05.0

)20(5.0)2(

410

5055)55(

=−=

<<−=>=

−>=>

Z P Z P

Z P X P

(c) )5540( << X P



2

5328..01915.03413..0

)5.00()10(

)5.01(

10

5055

10

5040)5540(

=+=

<<+<<=

<<−=

−<<

−=<<

Z P Z P

Z P

Z P X P

(d) )5540( << X P

9772.04772.05...0

)20()40(

)24(

410

5055

410

5040)5540(

=+=

<<+<<=

<<−=

−<<

−=<<

Z P Z P

Z P

Z P X P



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2. A pet food manufacturer produces cans of cat food with a nominal content

weight of 400 grams per can, however the can-filling machine yields acontent weight standard deviation of 20 grams. The cans are supplied to

wholesalers in boxes of 64 cans, and wholesalers require that the mean can

weight per box be at least 400 grams. To reduce the probability of a box of cat food not meeting a wholesaler’s requirements, the machine is set to

produce a mean can content weight of 403 grams. Calculate the probability

that a randomly selected box of cat food does not yield a mean can weightof at least 400 grams.

Let = X weight of can in grams then X ∼ ))20(,403?(2

Since n=64 is large by the central limit theorem

X ∼

64

)20(,403

2

N approximately & hence

1151.03849.05.0

)2.10(5.0

)2.1(

)820(

403400)400(

=−=

<<−=

−<=

−<=<

Z P

Z P

Z P X P

3. Recall the Anzac Garage data used in Week 3, Question 7. These data are

available from the course website in an EXCEL file called ANZACG.XLS.Use these 117 observations on used passenger cars to find the 95%

confidence interval for the population mean distance traveled by used

passenger cars (variable is denoted by odometer in the data set and is

measured in kilometers). Assume the population standard deviation is60,000kms.

Since n=117 is large we invoke the central limit theorem

X ~

117

000,60,

2

µ N at least approximately.

Using EXCEL we find the sample mean is 78,561 kms

The 95% confidence interval is given by



4

)433,89,689,67(

872,10561,78

117

000,6096.1561,78

025.0

=

±=

±=±n

z xσ

The calculated interval is one of the possible realizations of the 95%

confidence interval. In repeated sampling, 95% of intervals calculated in this

way would contain the true µ .

4. What would be the effects on the width of the confidence interval

calculated in Question 5 above of:(a) a decrease in the level of confidence used?

Decreases width (b) an increase in sample size?

Decreases width (c) an increase in the population standard deviation?

Increases width(d) an increase in the sample standard deviation?

No effect on the width since we are told the population standard deviation.

(e) an increase in the value of x found?

No effect on the width

5. Again referring to the statistical population in Question 3 above, determine

the sample size required to estimate the population mean to within 5,000kms with 90% confidence.

000,60,000,5,645.105.02 /

==== σ α

B z z

Sample size required

67.389000,5

)000,60(645.122

2 / =

=

=

B

zn

σ α

A sample size of 390 would be required.