12 distributions

14-04-2012

1

-4 -3 -2 -1 0 1 2 3 4

Research MethodologyDr. Nimit Chowdhary

Let X be a random variable. If the number of possible values of X is finite, or countably infinite, X is called a discrete random variable.

A distribution of this variable is called discrete distribution.

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2

In a simple toss of two coins the possible out comes are:

Head, Head (H, H)

Head, Tail (H, T)

Tail, Head (T, H)

Tail, Tail (T, T)

The distribution of “Heads”:

2 Heads 1 possibility

1 Head 2 possibilities

0 Head 1 possibility

Total: 4 Possibilities

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Probability distribution of heads

P(0 Heads) ¼ = 0.25

P(1 Head) ½ = 0.50

P(2 Heads) ¼ = 0.25

Total 1/1 = 1.00

Can you plot this information?

A sample of 100 Sandwiches prepared in a morning in a restaurant were examined and the number of imperfections per sandwich were counted with the following result.

Imperfections 0 1 2 3 4 5

Sandwiches 30 35 15 10 6 4

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4

The discrete probability distribution is given by

xi 0 1 2 3 4 5

P(xi) 30/100 35/100 15/100 10/100 6/100 4/100

00.05

0.10.15

0.20.25

0.30.35

0.4

1 2 3 4 5 6

When population is infinite, from a continuous stream

Random sample is taken without replacement Dichotomous outcomes- pass/fail,

accepted/rejected, heads/tails Process quality is constant

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A random sample of 5 tourist is selected from a tourists visiting a resort in Orcha, and the proportion dissatisfied is 0.10. What is the probability of 1 dissatisfied tourist in a sample of 5? What is the probability of 1 or less? What is the probability of 2 or more?

dndd

n qpCdP ..)(P(d) = Probability of d nonconforming unitsn = number in the sampled = number non-conforming in the

samplep = proportion nonconforming in the

populationq = proportion conforming (1-p) in the

population

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p = 0.10, q = 1-0.10 = 0.90n = 5, d = 1

328.0)1(

)90.0.()10.0.()!15(!1

!5)1(

.)!(!

!)(

41

P

P

qpdnd

ndP dnd

590.0)0(

)90.0.()10.0.()!05(!0

!5)0( 41

P

P

Now, the probability of 1 or less non dissatisfied tourist is P(0) + P(1)

So, P(1 or less) = P(0) + P(1)

= 0.590 + 0.328= 0.918

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Now, the probability of 2 or more dissatisfied tourists is: 1 - P(1 or less)

P(2 or more) = 1.00 – 0.918

= 0.082

When population is infinite Open ended situations When the process average (generally a

rate) is known n is quite large p is very small

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The average count of billing errors at a local restaurant per 8-hr shift is 1.0. What is the probability of 2 billing errors? The probability of 1 or less? The probability of 2 or more?

,!

)( ec

cPc

P(c) = Probability of c countsc = count or number = average count or average number e = 2.718281

np

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= 1.00, c = 2

184.0)2(!2

1!

)( 12

P

eec

cPc

Similarly,

368.0!1

1)1(

368.0!0

1)0(

11

10

eP

eP

P(1 or less) = P(0) + P(1)

= 0.368 + 0.368

= 0.736

P(2 or more) = 1.000 – P(1 or less)

= 1.000 – 0.736

= 0.264

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The number of deficiencies on a burger which is prepared on an automated assembly line has been found to follow the Poisson distribution with = 3. What is the probability that a randomly selected Burger will have two or less deficiencies.

32

0.

!3)2(

e

ccP

c

c

224.0.!2

3)2(

150.0.!1

3)1(

049.0.!0

3)0(

32

31

30

eP

eP

eP

423.0224.0150.0049.0)2( cP

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11

© Dr. Nimit ChowdharyResearch Methodology Workshop p.

21 Saturday, April 14, 2012

Also called Gaussian Distribution Most important in quality

methods A continuous distribution

© Dr. Nimit Chowdhary Research Methodology Workshop p. 22

Saturday, April

14, 2012

22

22

3989.021)(

zz

eeZf

-4 -3 -2 -1 0 1 2 3 4

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12

© Dr. Nimit ChowdharyResearch Methodology Workshop p.

23 Saturday, April 14, 2012

Bell Shaped Symmetrical Unimodal Mean = Median = Mode Area under the curve = 1 Asymptotic


Normal curve with different means but identical standard deviations

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Saturday, April

14, 2012

Normal curve with different standard deviations but identical means


Saturday, April

14, 2012

Percent of items included between certain values of the standard deviation

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Saturday, April

14, 2012

iXZ


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Saturday, April

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Saturday, April

14, 2012

The mean value of the weight of a particular brand of cereal for the past year is 0.297 kg (10.5 Oz.) with a standard deviation of 0.024 kg. Assuming a normal distribution, find the percent of data that falls below the specification limit of 0.274 kg.

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Saturday, April

14, 2012

96.0024.0

297.0274.0

Z

Z

XZ i

From table A it is found that for Z = -0.96, Area = 0.1685 or 16.85%.

Thus, 16.85 % of the data are less than 0.274 kg


Saturday, April

14, 2012

Determine the percentage of data that falls above 0.347 kg

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17

08.2024.0

297.0347.0

2

2

2

Z

Z

XZ i

From table A it is found that for Z2 = 2-08, Area2= 0.9812

Area1= Area T – Area2

= 1.000 – 0.9812 = 0.0188 or 1.88%

Thus, 1.88 % of the data are above 0.347 kg


Saturday, April

14, 2012

A large sample of Auto charges for a visit to Fort show a mean of Rs.118.5 and a population standard deviation of Rs.1.2. Determine the percentage of data between Rs. 116 and Rs.120

Auto drivers tend to bargain around the price set by Traffic Police.

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Saturday, April

14, 2012

08.220.1

5.118116

2

2

Z

Z

25.120.1

5.118120

3

3

Z

Z


Saturday, April

14, 2012

Z2 = -2.08, A2 = 0.0188, for Z3 = 1.25, A3 = 0.8944

So,

Area 1 = Area 3 – Area 2= 0.8944 – 0. 0188

= 0.8756

Thus, 87.56% of the data are between Rs.116 and Rs.120 .

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Saturday, April

14, 2012

If it is desired to have 12.1% of the Auto charges below Rs.115, how should the Traffic Police adjust the prices? The dispersion is = Rs.1.20.


Saturday, April

14, 2012

VX

X

XXZ i

4.11620.1

11517.1

0

0

01

Thus, the mean price should be centered at Rs. 116.4 for 12.1% of the values to be less than Rs. 115.

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Saturday, April

14, 2012

A cold-cereal manufacturer wants only 1.5% of the product to be below the weight specification of 0.567 kg (1.25 lb). If the weights are normally distributed and the standard deviation of the cereal filling machine is 0.018 kg, what mean weight is required?


Saturday, April

14, 2012

0.567 kg =X kg

1.5 %

= 0.018 kg

Z corresponding to 1.5 % area on left is – 2.17

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Saturday, April

14, 2012

kg

XZ i

60606.018.0

567.017.2


Saturday, April

14, 2012

The population mean of a company’s racing bicycles is 9.07 kg (20.0lb) with a population standard deviation of 0.40 kg. If the distribution is approximately normal, determine (a) the percentage of bicycles less than 8.30 kg, (b) the percentage of bicycles greater than 10.00 kg, and (c) the percentage of bicycles between 8.00 and 10.10 kg

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Saturday, April

14, 2012

= 9.07 kg = 0.40 kg

Less than8.30 kg


Saturday, April

14, 2012

925.140.0

07.930.8

Z

ZXZ i

For Z = - 1.925, Area = 0.0274 or 2.74 % cycles

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Saturday, April

14, 2012

= 9.07 kg = 0.40 kg

More than10.00 kg


Saturday, April

14, 2012

33.240.0

07.900.10

Z

ZXZ i

For Z = 2.33, Area = 0.9901

That means (1-0.9901) or 0.0099 or about 1% cycles

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Saturday, April

14, 2012

= 9.07 kg = 0.40 kg

8.00 kg 10.10 kg

Between


Saturday, April

14, 2012

675.240.0

07.900.80.80.8

ZXZ i

For Z =-2.67, Area = 0.0038

575.240.0

07.910.1010.1010.10

ZXZ i

For Z =2.575, Area = 0.9949Therefore, total cycles in this range

9911.00038.09949.00.810.10 AA

Or 99.11 % of bicycles produced

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Saturday, April

14, 2012

A large retail organisation has recently been receiving a number of complaints from customers about one of its products- bottles of own-brand shampoo. Customers have complained that they think the actual contents of the shampoo bottles are some times less than they are meant to be according to the printed contents label on each bottle. The organisation currently sells two bottle sizes: one at 490 m.l. And one at 740 m.l.

The shampoo is actually manufactured and bottled by a subsidiary company and an investigation into the bottling process is currently underway. The process is such that two production lines are in use- one for each bottle size.


On each line a computer-controlled machine fills the bottle with shampoo automatically. However management realizes that the machine are not 100% accurate and some variability inevitably arises in the exact volume of shampoo put into each bottle. Further analyses revealed that on one line the machine is calibrated to deliver 500 ml of shampoo into each bottle but that the actual process in normally distributed with a standard deviation of 10 ml.

On the second line a similar situation applies with the machine set to deliver mean amount of 750 ml with the standard deviation of 15 ml and again the distribution amount filled is normally distributed.

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Activity 1

Calculate the proportion of bottles that contain less than 490 ml.


Saturday, April

14, 2012

Activity 2

As such we can not continue with almost 16% of output below the advertised weight.

While the company has decided that machine 1 will continue as before (mean 500 ml and standard deviation 10 ml) but that a new label will be produced showing the new minimum contents and guaranteeing that no more than 1% of the output will fall below this minimum figure.

Determine this new figure.

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Saturday, April

14, 2012

Activity 1

%161587.0

0.110

500490

Z

Activity 2

For 1%, i.e. Corresponding to .01 Z= -2.33

10500

..33.2

X

dsX

Or –23.3= X-500

Or X= 476.7 ml

12 distributions

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