Outline Solutions to Tutorial Sheet 1

Graham Crocker

Description

Outline Solutions to Tutorial Sheet 31. Not true! The vast majority of people have both eyes the same colour so if the left eye isbrown the right eye will almost certainly also be brown. So the probability that both eyes arebrown is also about 0.6.More formally, the events are not independent soP(left eye brown and right eye brown) ≈P(left eye brown)×P(right eye brown given left eye brown) = 0.6×1 = 0.62. From the tree diagram:JointProbabilityP( MachineA ∩ Defective ) = 0.45×0.08 = 0.036P( ∩ NotDefective ) = 0.414×0.92P( MachineB ∩ Defective) = 0.055= 0.55×0.10P( ∩ NotDefective ) = 0.495×0.90TOTAL = 1.000Prob(item from Machine A given it’s defective)P(A ∩ Defective) 0.036= P(A | Defective) = = = 0.396P() 0.036 + 0.0553. Let X be the number of defective bulbs found.6 5 4 120(i) P(X = 0) = × × = = 0.2389 8 7 5043 6 5 6 3 5 6 5 3 (ii) P(X = 1) = × × + × × + × × = 3× 0.1786 = 0.536 9 8 7 9 8 7 9 8 7 (iii) P(X ≥ 1) = 1− P(X = 0) = 1− 0.238 = 0.7624. It helps to draw a tree diagram first. Then:P(D ∩ +) 0.05× 0.95P(Has disorder | + result) = =P(+)()0.05× 0.95 +(0.95× 0.10)0.0475= = 0.3330.1425P(NotD ∩ −) 0.95× 0.90P(No disorder | – result) = =P(−) ()0.05× 0.05 +(0.95× 0.90)0.855= = 0.9970.8575If you have a negative result, you are almost certainly free of the disorder.If you have a positive result, you are still probably disorder free though your chances of havingit are increased. Without the test ...

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• Publié par: Mivug
• Nombre de lectures: 15
• Langue: English

Extrait

Outline Solutions to Tutorial Sheet 3

1.

Not true! The vast majority of people have both eyes the same colour so if the left eye is

brown the right eye will almost certainly also be brown. So the probability that both eyes are

brown is also about 0.6.

More formally, the events are

not independent

so

P(left eye brown

and

right eye brown)

≈

P(left eye brown)

×

P(right eye brown

given

left eye brown) = 0.6

×

1 = 0.6

2.

From the tree diagram:

Joint

Probability

P(

Defective

MachineA

∩

)

=

0.45

×

0.08

=

0.036

P(

ve

NotDefecti

MachineA

∩

)

=

0.45

×

0.92

=

0.414

P(

Defective

MachineB

∩

)

=

0.55

×

0.10

=

0.055

P(

ve

NotDefecti

MachineB

∩

)

=

0.55

×

0.90

=

0.495

TOTAL

=

1.000

Prob(item from Machine A

given

it’s defective)

396

.

0

055

.

0

036

.

0

036

.

0

)

(

)

(

)

|

(

=

+

=

∩

=

=

Defective

P

Defective

A

P

Defective

A

P

3.

Let

X

be the number of defective bulbs found.

(i)

238

.

0

504

120

7

4

8

5

9

6

)

0

(

=

=

×

×

=

=

X

P

(ii)

536

.

0

1786

.

0

3

7

3

8

5

9

6

7

5

8

3

9

6

7

5

8

6

9

3

)

1

(

=

×

=

÷

ø

ö

ç

è

æ

×

×

+

÷

ø

ö

ç

è

æ

×

×

+

÷

ø

ö

ç

è

æ

×

×

=

=

X

P

(iii)

762

.

0

238

.

0

1

)

0

(

1

)

1

(

=

−

=

=

−

=

≥

X

P

X

P

4.

It helps to draw a tree diagram first. Then:

P(Has disorder | + result)

(

)

(

)

10

.

0

95

.

0

95

.

0

05

.

0

95

.

0

05

.

0

)

(

)

(

×

+

×

×

=

+

+

∩

=

P

D

P

333

.

0

1425

.

0

0475

.

0

=

=

P(No disorder | – result)

(

)

(

)

90

.

0

95

.

0

05

.

0

05

.

0

90

.

0

95

.

0

)

(

)

(

×

+

×

×

=

−

−

∩

=

P

NotD

P

997

.

0

8575

.

0

855

.

0

=

=

If you have a negative result, you are almost certainly free of the disorder.

If you have a positive result, you are still probably disorder free though your chances of having

it are increased. Without the test you have a 5% chance of having it, with a positive test result

you have a 33% chance of having it.

5.

Part A

Both the normal distribution and the lognormal distribution are good models. In modelling

generally, we would use the simplest model which is acceptable. In this case the

normal

distribution.

Part B

The

exponential distribution

can be used. Note that the

Weibull

plot is also linear. As we will

see in a later section, the exponential distribution is a special case of the Weibull so this is as

expected.

7000

6000

5000

4000

3000

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

Percent

Normal Probability Plot for Part A

ML Estimates

Mean:

StDev:

4944.11

651.096

7000

6000

5000

4000

3000

2000

1000

0

99

98

97

95

90

80

70

60

50

30

10

Data

Percent

Exponential Probability Plot for Part B

ML Estimates

Mean:

861.142

Part C

The

Weibull distribution

does best in the tails of the ID plot and all points are within limits in

the Weibull plot below.

Part D

ID plots of all the data show a clear outlier which needs investigating. Looking at the data, this

is item 89 recorded as 22250.0 Why is it so extreme? If it’s a

correct

value, we would want to

know why this item lasted so long – which would probably be the most interesting piece of

information in the data. If it’s

incorrect

, the value should be corrected or

removed. Without

any other information, it’s best to just remove it. (Actually, it was a data entry error. 22250.0

was entered instead of 2250.0). Plots with the outlier removed show that the

Weibull

distribution

is an excellent fit.

2000

1500

1000

99

95

90

80

70

60

50

40

30

20

10

5

3

2

1

Data

Percent

Weibull Probability Plot for Part C

ML Estimates

Shape:

Scale:

9.55028

1959.22

1000

100

99

95

90

80

70

60

50

40

30

20

10

5

3

2

1

Data

Percent

Weibull Probability Plot for Part D

ML Estimates

Shape:

Scale:

2.31695

940.548

Part E

The

lognormal distribution

is a good model.

Part F

ID plots (except the exponential) are S-shaped which suggests the distribution has more than

one peak. A histogram would help here. In fact it’s

bimodal

.

10000

1000

100

10

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

Percent

Lognormal Probability Plot for Part E

ML Estimates

Location:

Scale:

5.06924

1.02544

4000

3000

2000

20

10

0

Part F

Frequency

Also on the worksheet in column C43 is an indicator for

Part F Type

. This shows that the first

50 items in column 6 are of one type and the second 50 are of another type. These should be

analysed separately. Splitting (stratifying) by type suggests that the original distribution may be

considered as a mixture of normal distributions. The probability plots are given below.

6.

(i) P(lifetime < 6000) = 0.95 approximately

(ii) 90% of items have lifetimes less than 5800 hours approximately which is the same as

saying that 10% of items will have lifetimes exceeding this amount.

7.

(i) P(lifetime > 3000) = 1 – p(lifetime < 3000) = 1 – 0.97 approximately. Thus this item has

3% reliability at 3000 hours.

(ii) The median will have 50% of lifetimes below it. From the plot this is roughly 550 hours.

Using Minitab, the mean of the sample of 100 lifetimes is 861 hours. From the plot just over

60% (or a proportion of 0.60) will fail before this time. The distribution is very positively skew

and, as expected, the mean is larger than the median.

8.

P(1500 < lifetime < 2000) = P(lifetime < 2000) – P(lifetime < 1500)

From the plot this is approximately 0.70 – 0.075 = 0.625

9.

The reliability at 1000 hours = P(lifetime > 1000)

≈

1 – 0.96 = 0.04

Looking at the data in column 5 of the worksheet, it can be seen that 7 of the 100 items had

lifetimes of at least 1000 hours – a relative frequency of 7/100 = 0.07. This is similar to the

estimate obtained from the probability plot. However, because the estimate from the plot uses

more of the information in the data, it is likely to be closer to the true population value.

1

2

4000

3000

2000

99

95

90

80

70

60

50

40

30

20

10

5

1

Data

Percent

Normal Probability Plot for Part F By Part F Type