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Chapter 4. 4-1. Discrete Probability Distributions. Outline. 4-2. 4-1 Introduction 4-2 Probability Distributions and Expectation 4-3 The Binomial Distribution 4-4 Mean, Variance and Standard Deviation for the Binomial Distribution. Objectives. 4-3.

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Chapter 4

Chapter 4

4-1

Discrete Probability Distributions


Outline
Outline

4-2

  • 4-1 Introduction

  • 4-2 Probability Distributions and Expectation

  • 4-3 The Binomial Distribution

  • 4-4 Mean, Variance and Standard Deviation for the Binomial Distribution


Objectives
Objectives

4-3

  • Construct a probability distribution for a random variable.

  • Find the mean and expected value for a discrete random variable.

  • Find the exact probability forXsuccesses inntrials of a binomial experiment.


4 2 probability distributions
4-2 Probability Distributions

4-5

  • A variable is defined as a characteristic or attribute that can assume different values.

  • A variable whose values are determined by chance is called a random variable.


4 2 probability distributions1
4-2 Probability Distributions

4-6

  • If avariablecan assume only a specific number of values, such as the outcomes for the roll of a die or the outcomes for the toss of a coin, then the variable is called adiscrete variable.

  • Discrete variables have values that can be counted.


4 2 probability distributions2
4-2 Probability Distributions

4-7

  • If avariablecan assume all values in the interval between two given values then the variable is called acontinuous variable.Example -temperature between 680 to 780.

  • Continuous random variables are obtained from data that can be measured rather than counted.


4 2 probability distributions tossing two coins

H

H

T

Second Toss

H

T

4-2 Probability Distributions - Tossing Two Coins

4-8

T

First Toss


4 2 probability distributions tossing two coins1
4-2 Probability Distributions - Tossing Two Coins

4-9

  • From the tree diagram, the sample space will be represented byHH, HT, TH, TT.

  • IfXis the random variable for the number of heads, thenXassumes the value0, 1, or 2.


4 2 probability distributions tossing two coins2
4-2 Probability Distributions - Tossing Two Coins

4-10

Sample Space

Number of Heads

0

1

2

TT

TH

HT

HH



4 2 probability distributions3
4-2 Probability Distributions

4-12

  • Aprobability distributionconsists of the values a random variable can assume and the corresponding probabilities of the values. The probabilities are determined theoretically or by observation.


4 2 probability distributions graphical representation
4-2 Probability Distributions -- Graphical Representation

4-13

Experiment: Toss Two Coins

1

Y

T

I

L

I

0

.

5

B

A

B

O

R

P

.25

0

1

2

3

N

U

M

B

E

R

O

F

H

E

A

D

S


4 2 mean variance and expectation for discrete variable

The

mean

of

the

random

variable

of

a

probability

distribution

is

m

×

+

×

+

+

×

=

X

P

(

X

)

X

P

(

X

)

.

.

.

X

P

(

X

)

1

1

2

2

n

n

=

×

å

X

P

(

X

)

where

X

,

X

,

.

.

.

,

X

are

the

outcomes

and

1

2

n

P

(

X

),

P

(

X

),

.

.

.

,

P

(

X

)

are

the

corresponding

1

2

n

probabilities.

4-2 Mean, Variance and Expectation for Discrete Variable

4-14


4 2 mean for discrete variable example
4-2 Mean for Discrete Variable -Example

4-15

  • Find the mean of the number of spots that appear when a die is tossed. The probability distribution is given below.


4 2 mean for discrete variable example1
4-2 Mean for Discrete Variable -Example

4-16

That is, when a die is tossed many times,

the theoretical mean will be 3.5.


4 2 mean for discrete variable example2
4-2 Mean for Discrete Variable -Example

4-17

  • In a family with two children, find the mean number of children who will be girls. The probability distribution is given below.


4 2 mean for discrete variable example3
4-2 Mean for Discrete Variable -Example

4-18

X

P

(

X

)

0

(

1

/

4

)

1

(

1

/

2

)

2

(

1

/

4

)

=

1

.

That is, the average number of

girls in a two-child family is 1.



4 2 expectation example
4-2 Expectation -Example

4-26

  • A ski resort loses $70 000 per season when it does not snow very much and makes $250 000 when it snows a lot. The probability of it snowing at least 190 cm (i.e., a good season) is 40%. Find the expected profit.


4 2 expectation example1
4-2 Expectation -Example

4-27

  • The expected profit = ($250 000)(0.40) + (–$70 000)(0.60) = $58 000.


4 3 binomial probability distributions
4-3 Binomial Probability Distributions

4-28

  • Abinomial experimentis a probability experiment that satisfies the following four requirements:

  • Each trial can have only two outcomes or outcomes that can be reduced to two outcomes. Each outcome can be considered as either a success or a failure.


4 3 the binomial distribution
4-3 The Binomial Distribution

4-29

  • There must be a fixed number of trials.

  • The outcomes of each trial must be independent of each other.

  • The probability of success must remain the same for each trial.


4 3 the binomial distribution1
4-3 The Binomial Distribution

4-30

  • The outcomes of a binomial experiment and the corresponding probabilities of these outcomes are called abinomial distribution.


4 3 the binomial distribution2
4-3 The Binomial Distribution

4-31

  • Notation for the Binomial Distribution:

  • P(S) = p, probability of a success

  • P(F) = 1 – p = q, probability of a failure

  • n= number of trials

  • X=number of successes.


4 3 binomial probability formula
4-3 Binomial Probability Formula

4-32

In

a

binomial

experiment

,

the

probability

of

exactly

X

successes

in

n

trials

is

n

!

P

(

X

)

p

q

X

n



X

(

n

X

)

!

X

!


4 3 binomial probability example
4-3 Binomial Probability -Example

4-33

  • If a student randomly guesses at five multiple-choice questions, find the probability that the student gets exactly three correct. Each question has five possible choices.

  • Solution:n = 5, X = 3 and p = 1/5. Then, P(3) = [5!/((5–3)!3! )](1/5)3(4/5)2 =0.0512.


4 3 binomial probability example1
4-3 Binomial Probability -Example

4-34

  • A survey found that 30% of teenage consumers received their spending money from part-time jobs. If five teenagers are selected at random, find the probability that at least three of them will have part-time jobs.


4 3 binomial probability example2
4-3 Binomial Probability -Example

4-35

  • Solution:n = 5, X = 3, 4 and 5 and p = 0.3. Then, P(X3) = P(3) + P(4) + P(5) = 0.1323 + 0.0284 + 0.0024 = 0.1631.

  • NOTE:You can use Table B in the textbook to find the Binomial probabilities as well.


4 3 binomial probability example3
4-3 Binomial Probability -Example

4-36

  • A report from the Canadian Safety Commission stated that 70% of single-vehicle traffic fatalities that occur on weekend nights involve an intoxicated driver. If a sample of 15 single-vehicle traffic fatalities that occurred on a weekend night is selected, find the probability that exactly 12 involve a driver who is intoxicated.


4 3 binomial probability example4
4-3 Binomial Probability -Example

4-37

  • Solution:n = 15, X = 12 and p = 0.7. From Table B, P(X =12) = 0.170


4 4 mean variance standard deviation for the binomial distribution example
4-4 Mean, Variance, Standard Deviation for the Binomial Distribution -Example

4-38

  • A coin is tossed four times. Find the mean, variance and standard deviation of the number of heads that will be obtained.

  • Solution:n = 4, p = 1/2 and q = 1/2.

  •  = np = (4)(1/2) = 2.

  • 2 = npq = (4)(1/2)(1/2) = 1.

  •  = = 1.