TU QM -L7 Discrete Probability_updated

7/27/2019 TU QM -L7 Discrete Probability_updated

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LECTURE 7PROBABILITY DISTRIBUTION

1

Discrete Probability Distribution

Binomial DistributionPoisson Distribution


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Introduction to Probability Distributions

Random Variable

Represents a possible numerical value from arandom event

Takes on different values based on chance

Random

Variables

DiscreteRandom Variable

ContinuousRandom Variabletoday

Nextweek

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A discrete random variable is a variable that is

determined by counting ie can assume only a

countablenumberof values

Many possible outcomes:

number of complaints per day

number of TVs in a household

number of rings before the phone is answered

Only two possible outcomes: gender: male or female

defective: yes or no

spreads peanut butter first vs. spreads jelly first

Discrete Random Variable

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I. The sum of the probabilities of all the

outcomes is 1. P(X) = 1II. The probability of a particular

outcome is between 0 and 1.

0

P(X)

1III. The outcomes are mutually

exclusive.

Main features of

Discrete Probability Distributions


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Mean of a probability distribution

Variance of a probability distribution

Standard Deviation of a probability distribution

2

2

222

)()(

)()()(

xxPxPx

XEXEXVar

The Mean, Variance and Standard

Deviat

ion of a Discrete Probability

Distributions

)()( xxPXE

)()( XVarXSD


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EXAMPLE

Clifton Windows and Glass Company makes and distributes

window products for new home constructions. Each week the

companys quality manager examines a randomly selected window tocheck for defects

No of

Defects, X Frequency

0 1501 110

2 50

3 90

400

1) Find probability for each defect.

2) Calculate the Mean: Expected value, E(x)

3) Calculate the Standard Deviation, 6


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E(x) = xP(x) = ??

1) The Mean: Expected value, E(x)

7

No of

defects,

x

Frequency P(x) xP(x) xP(x)

01

2

3

150110

50

90

400


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2) Standard Deviation,

22 [E(x)]-)E(x

8


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LAWS OF EXPECTED VALUE AND VARIANCE

EXPECTED VALUE ; E(X)

1. E(c) = c

2. E(cX) = c x E(X)

VARIANCE, VAR(X)

1. Var(c) = 0

2. Var (c X) = c Var(X)


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EXAMPLE:

Discrete random variable X is given by P(X = x) = cX

For x = 1, 2, 3, 4.Construct probability distribution table and compute the

Mean value.


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Probability Distributions

Continuous

Probability

Distributions

Binomial

Poisson

Probability

Distributions

Discrete

Probability

Distributions

Normal

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Continuous

Probability

Distributions


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The Binomial Distribution

Characteristics of the Binomial Distribution:

A trial has only two possible outcomessuccess or

failureThere is a fixed number, n, ofidentical trials

The trials of the experiment are independentof each

other

Theprobability of a success, p, remains constantfrom trial to trial

If p represents the probability of a success, then

(1-p) = q is the probability of a failure


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Binomial Distribution Settings

A manufacturing plant labels items as

either defective or acceptable

A firm bidding for a contract will either getthe contract or not

A marketing research firm receives survey

responses of yes I will buy or no I willnot

New job applicants either accept the offer

or reject it


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Counting Rule for Combinations

A combination is an outcome of an experiment

where x objects are selected from a group of n

objects

)!xn(!x!nCnx

where:

Cx = number of combinations of x objects selected from n objectsn! =n(n - 1)(n - 2) . . . (2)(1)

x! = x(x - 1)(x - 2) . . . (2)(1)

0! = 1 (by definition)

n


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P(x) = nCx p^x q^x (using calculator)

P(x) = probability ofx successes in n trials,

with probability of success pon each trial

x = number of successes in sample,(x = 0, 1, 2, ..., n)

p = probability of success per trial

q = probability of failure = (1 p)

n = number of trials (sample size)

P(x) p qx n x

Example: Flip a coin four

times, let x = # heads:

n = 4

p = 0.5

q = (1 - .5) = .5

x = 0, 1, 2, 3, 4

Binomial Distribution Formula

n

xC


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Binomial Distribution Characteristics

Mean

Variance and Standard Deviation

npE(x)

npq2

npq Where n = sample size

p = probability of success

q = (1 p) = probability of failure


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Binomial Distribution ExampleExample: 35% of all voters support

Proposition A. If a random sample of 10voters is polled, what is the probability that

exactly three of them support the

proposition?

i.e., find P(x = 3) if n = 10 and p =0.35


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A survey shows that 30% of students major inBusiness Administration. Consider a random

sample of 10 students, find the probability that

the number of students major in Business

Administration would be:a) Three

b) None

c) Less than twod) At least one

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Example of a Binomial Distribution


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If X P ( ), then

where

is the mean number of successes ina particular interval

e is the constant 2.71828

x is the number of successes P(x) is the probability for a specified value

of x

mean = E(X) = variance = V(X) =

x!ex)P(X

x

Poisson Distributions Formula


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In a Poisson distribution mean, = 0.3. Find:

(a) P(x= 0)

(b) P(x 1)

P(X 1) = P(X=1) + P(X=2) + (until infinity)= 1 [P(X=0)]

EXAMPLE


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The marketing manager of a company has noted that she

usually receives 10complaint calls from customers during

a working week (A working week is a 5 day week) and

these calls occur at random.

Find the probability of her receiving more than one call in

a single day.

Solution:X = the number of calls per weekX P (10)Y = the number of calls per day

Y Poisson (10/5 = 2)P( Y > 1 ) = P (Y=2) + P(Y=3) +

= 1 [P(Y=0) + P(Y=1)]

594.0

22

1!

2e

0!

2e-1

10

EXAMPLE


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QUESTION

Each day a quality control inspector selects 10 items

from a continuous production line. Fromexperience, he knows that 30% of the items

produced on this line will have to be modified.

(a) What is the probability that out of 10 samples

taken, none of them need any modification ?

(b) What is the probability that out of 10 samples,

there will be at least 3 that need modification

(c) On average, how many need modification out of15 samples?

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TU QM -L7 Discrete Probability_updated