2 a Probability Distributions

Probability & Distributions

Electronics Test & Development Centre, Chennai

Ministry of Communications & Information Technology



Random Experiment

An experiment whose outcome is not predictable in advance,

but all possible outcomes are known.

Sample space:

The set of all possible outcomes is called a sample

space,denoted by S.

Eg. (1) Random experiment : Tossing two fair coins

S = {HH, HT, TH, TT}

(3) Life time of the bulb : S = {0, }

Here Life Testing is an Random Experiment

Event : Any subset of the sample space is called an event,

generally denoted by A, B,….

Probability Concepts





Probability Axioms

(1) First axiom:

The probability of an event is a non-negative real number:

(2) Second axiom : The probability of the Sample space =1

(3) Third axiom :

Any countable sequence of disjoint events E1,E2,... satisfies

When the events are not disjoint then:





http://en.wikipedia.org/wiki/Countable

• If there are 35 black and 65 white marbles are there

in a black box. The probability that one marble

selected at random from the box is black is 0.35.

This could be determined by the following.

1. Set the TOTAL to 0

2. Select one marble at random from the box.

3. Determine the marble’s color. Add +1 to our total if the

marble is black, 0 otherwise.

4. Repeat steps 2 and 3 at least 1000 times, each time

selecting from all 100 marbles (simple random selection

with replacement).

5. Divide the TOTAL observed by the total number of iterations

(1000) and that fraction would be the estimated probability

of selecting a black marble at random.

Probability = Relative frequency

1. What is the probability of observing 3 or 4 on the toss

of a single die?

2. What is the probability of observing a total of 13 on

the toss of two dies?

3. What is the probability of observing a total of 5 on the

toss of two dies?

4. A coin and a die are tossed simultaneously. What is the

probability of getting tail from coin tossing and

observing even number on a Die.

5. In a box there are 6 blue and 4 red marbles are there.

What is the probability that one marble selected at

random from the box is red?

Exercise on Probability





• Binomial Distribution

• Poisson Distribution



Discrete Distributions

• The experiment consists of n identical trials (simple experiments).

• Each trial results in one of two outcomes (success or failure)

• The probability of success on a single trial is equal to p and p remains

the same from trial to trial.

• The trials are independent, that is, the outcome of one trial does not

influence the outcome of any other trial.

• The random variable y is the number of successes observed during n

trials.

Mean

Standard deviation = np

=



Binomial Distribution

Experiment: 5 fair coins are tossed.

Event of interest: the number of heads.

Each coin represents a Bernouilli Trial

with probability of a head coming up (a

success) of .5.

The sum of Bernouilli trials is a binomial

random variable, in this case with n=5,

p=.5.

Class Frequency

0 1

1 11

2 11

3 19

4 6

5 2

The Experiment is repeated 50 times.

0 1 2 3 4 5

20

18

16

14

12

10

8

6

4

2

0



Binomial Distribution Example

Binomial Distribution - THeoritical Example

0.03125

0.15625

0.3125 0.3125

0.15625

0.03125

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0 1 2 3 4 5

No. of Heads

Pd

f valu

e f

or

gett

ing

no

. o

f

head

s

A random variable X is said to have a Poisson

Distribution if

P(X=x) = e- x / x ! x = 0,1,……. > 0

is called the parameter of the distribution.

Other Forms of Poisson Distribution:

1) P(X=x) = e-np (np)x / x !

where x no. of defects in iron sheet (or) fabric (or) no. of spelling mistakes

in a volumunised book etc. n = for a given value p = probability of getting

a defect (from the previous available data)

2) P(X=r) = e-t (t)r / r !

where r = the no. of failures in time t = failure rate per hour based

on the data

t = for a given time expressed in hours

P(r) = Probability of getting r failures in time t



Poisson Distribution

Example of Poisson Distribution-Wars by Year

• Number of wars beginning by year for years 1482-1939.

Table of Frequency counts and proportions (458 years):

wars Frequency Proportion

0 242 0.5284

1 148 0.3231

2 49 0.1070

3 15 0.0328

4 4 0.0087

More 0 0

• Total Wars: 0(242) + 1(148) + 2(49) + 3(15) + 4(4) = 307

• Average Wars per year: 307 wars / 458 years = 0.67 wars/year

Using Poisson Distribution as Approximation

• Since mean of empirical (observed) distribution

is 0.67, use that as mean for Poisson distribution

(that is, set = 0.67)

– p(0) = (e-0)/0! = e-0.67 = 0.5117

– p(1) = (e-1)/1! = e-0.67(0.67) = 0.3428

– p(2) = (e-2)/2! = e-0.67(0.67)2/2 = 0.1149

– p(3) = (e-3)/3! = e-0.67(0.67)3/6 = 0.0257

– p(4) = (e-4)/4! = e-0.67(0.67)4/24 = 0.0043

– P(Y5) = 1-P(Y4)=

1-.5117-.3428-.1149-.0257-.0043=0.0006

Comparison of Observed and Model

Probabilities

• In EXCEL, the function =POISSON(y,,FALSE)

returns p(y) = e-y/y!

wars Frequency Proportion Model

0 242 0.5284 0.5117

1 148 0.3231 0.3428

2 49 0.1070 0.1149

3 15 0.0328 0.0257

4 4 0.0087 0.0043

More 0 0 0.0006

The model provides a good fit to the observed data. Formal

tests of goodness-of-fit are covered in the sequel course.



Poisson Distribution – Empirical Data Graph

Poisson Distribution Empirical -Graph

(No. of Wars held during 1482-1939)

242

148

49

154 0 0 0 0 0 0

0

50

100

150

200

250

300

1 2 3 4 5 6 7 8 9 10 11No. of Wars

No

. o

f Y

ears



Poisson Distribution – Theoretical -Graph

Poisson PDF values (theoritical) for occuring

no. of warsin a year

0

0.1

0.2

0.3

0.4

0.5

0.6

0 1 2 3 4 5 6 7 8 9 10No. of wars

Pro

b o

f o

ccu

rin

g n

o. o

f w

ars

• Normal Distribution

• Exponential Distribution



Continuous Distributions











Find P(2 < X < 4) when X ~ N(5,2).

The standarization equation for X is:

Z = (X-)/ = (X-5)/2

when X=2, Z= -3/2 = -1.5

when X=4, Z= -1/2 = -0.5

P(2<X<4) = P(X<4) - P(X<2)

P(X<2) = P( Z< -1.5 )

= P( Z > 1.5 ) (by symmetry)

P(X<4) = P(Z < -0.5)

= P(Z > 0.5) (by symmetry)

P(2 < x < 4) = P(X<4)-P(X<2)

= P(Z>0.5) - P( Z > 1.5)

= 0.3085 - 0.0668 = 0.2417

x

y

-4-3-2-101234



• If a continuous variable is monitored (such as

the length of the rod from cutting process) the

variable will usually be distributed normally

about a mean .

• Spread of values may be measured in terms of

population standard deviation which defines

the width of the bell-shaped curve.

• = 150 mm

• = 5 mm

The Normal Distribution

68.27% of the steel rods produced will lie with in ± 5

mm of the mean ( ± )

95.45% of the rods will lie within ± 10 mm (( ± 2)..

99.73% of the rods will lie within ± 15 mm (( ± 3).





• From the table, at the value + 1.96

• 0.025 (2.5%) of the population exceeds this

length.

• Hence, 95% population lies within + 1.96.

• Similarly, 99.8% of the rod lengths lie with

+ 3.09



Lengths of 100 steel rods (mm) 144 146 154 146

151 150 134 153

145 139 143 152

154 146 152 148

157 153 155 157

157 150 145 147

149 144 137 155

141 147 149 155

158 150 149 156

145 148 152 154

151 150 154 153

155 145 152 148

152 146 152 142

144 160 150 149

150 146 148 157

147 144 148 149

155 150 153 148

157 148 149 153

153 155 149 151

155 142 150 150

146 156 148 160

152 147 158 154

143 156 151 151

151 152 157 149

154 140 157 151

Quality for

Competitiveness

August, 2003 MSQM-BITS, Pilani 22



Quality for

Competitiveness




Quality for

Competitiveness




Quality for

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Quality for

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=NORMDIST(260,255,50,TRUE)

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2 a Probability Distributions