6-1

Chapter 6

Probability
Distributions
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6-2 Outline
 6-1 Introduction

6-2 Probability Distributions

6-3 Mean, Variance, and
Expectation
 6-4 The Binomial Distribution

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6-3 Objectives

Construct a probability distribution for
a random variable.

Find the mean, variance, and expected
value for a discrete random variable.

Find the exact probability for X
successes in n trials of a binomial
experiment.

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6-4 Objectives

Find the mean, variance, and standard
deviation for the variable of a binomial
distribution.

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6-5 6-2 Probability Distributions

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.
variable

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6-6 6-2 Probability Distributions
 If a variable can 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 a discrete
variable.
variable
 Discrete variables have values that can
be counted.
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6-7 6-2 Probability Distributions
 If a variable can assume all values in
the interval between two given values
then the variable is called a continuous
variable. Example - temperature
between 680 to 780.
 Continuous random variables are
obtained from data that can be
measured rather than counted.
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 6-2 Probability Distributions -
Tossing Two Coins
6-8
H

H
T

Second Toss

T H

First Toss T
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 6-2 Probability Distributions -
6-9 Tossing Two Coins


From the tree diagram, the sample
space will be represented by HH,
HT, TH, TT.
 If X is the random variable for the

number of heads, then X assumes

the value 0, 1, or 2.
2

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 6-2 Probability Distributions -
6-10 Tossing Two Coins
Sample Space Number of Heads

TT 0

TH
1
HT

HH 2

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 6-2 Probability Distributions -
6-11 Tossing Two Coins

OUTCOME PROBABILITY
X P(X)
0 1/4

1 2/4

2 1/4

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6-12 6-2 Probability Distributions

 A probability distribution consists of

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

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 6-2 Probability Distributions --
6-13 Graphical Representation

Experiment: Toss Two Coins

1
PROBABILITY

0.5

.25

0 1 2 3

NUMBER OF HEADS

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 6-3 Mean, Variance, and
6-14 Expectation for Discrete Variable

The mean of the random variable of a

probabilitydistribution is
µ = X ⋅ P( X ) + X ⋅ P( X ) + ... + X ⋅ P( X )
1 1 2 2 n n

= ∑ 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 .

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 6-3 Mean for Discrete Variable -
6-15 Example

Find the mean of the number of spots
that appear when a die is tossed. The
probability distribution is given below.

X
X 11 22 33 44 55 66

P(X)
P(X) 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6 1/6
1/6

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 6-3 Mean for Discrete Variable -

6-16 Example

µ = ∑ X ⋅ P( X )
= 1 ⋅ (1 / 6) + 2 ⋅ (1 / 6) + 3 ⋅ (1 / 6) + 4 ⋅ (1 / 6)
+ 5 ⋅ (1 / 6) + 6 ⋅ (1 / 6)
= 21 / 6 = 35 .

That is, when a die is tossed many times,

the theoretical mean will be 3.5.
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 6-3 Mean for Discrete Variable -

6-17 Example

In a family with two children, find the
mean number of children who will be
girls. The probability distribution is
given below.

X
X 00 11 22

P(X)
P(X) 1/4
1/4 1/2
1/2 1/4
1/4

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 6-3 Mean for Discrete Variable -

6-18 Example

µ = ∑ 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.

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 6-3 Formula for the Variance of a
6-19 Probability Distribution

 The variance of a probability

distribution is found by multiplying the
square of each outcome by its
corresponding probability, summing
these products, and subtracting the
square of the mean.

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 6-3 Formula for the Variance of a
6-20 Probability Distribution
The formula for the variance of a
probability distribution is
σ = ∑ [ X ⋅ P( X )] − µ .
2 2 2

The standard deviation of a

probability distribution is
σ= σ . 2

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 6-3 Variance of a Probability
6-21 Distribution - Example

 The probability that 0, 1, 2, 3, or 4

people will be placed on hold when
they call a radio talk show with four
phone lines is shown in the
distribution below. Find the variance
and standard deviation for the data.

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 6-3 Variance of a Probability
6-22 Distribution - Example

X 0 1 2 3 4

P(X) 0.18 0.34 0.23 0.21 0.04

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 6-3 Variance of a Probability
6-23 Distribution - Example
X P(X) X⋅ P(X) X2⋅ P(X)
0 0.18 0 0 σ2 =
3.79 – 1.592
1 0.34 0.34 0.34
= 1.26
2 0.23 0.46 0.92
3 0.21 0.63 1.89
4 0.04 0.16 0.64
µ = 1.59 Σ X2⋅ P(X)
=3.79

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 6-3 Variance of a Probability
6-24 Distribution - Example
 Now, µ = (0)(0.18) + (1)(0.34) + (2)(0.23) +
(3)(0.21) + (4)(0.04) = 1.59.
2

Σ X P(X) = (02)(0.18) + (12)(0.34) + (22)(0.23)
+ (32)(0.21) + (42)(0.04) = 3.79

1.592 = 2.53 (rounded to two decimal
places).
 σ 2 = 3.79 – 2.53 = 1.26

σ = 1.26 = 1.12
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6-25 6-3 Expectation
The expected value of a discrete
random variable of a probability
distribution is the theoretical average
of the variable. The formula is
µ = E ( X ) = ∑ X ⋅ P( X )
The symbol E ( X ) is used for the
expected value.
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6-26 6-3 Expectation - Example
 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 75 inches (i.e., a good season)
is 40%. Find the expected profit.

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6-27 6-3 Expectation - Example
Profit, X 250,000 –70,000

P(X) 0.40 0.60

 The expected profit = ($250,000)(0.40)

+ (–$70,000)(0.60) = $58,000.

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6-28 6-4 The Binomial Distribution


A binomial experiment is 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.
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6-29 6-4 The Binomial Distribution
 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.

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6-30 6-4 The Binomial Distribution

The outcomes of a binomial
experiment and the corresponding
probabilities of these outcomes are
called a binomial distribution.

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6-31 6-4 The Binomial Distribution


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.
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6-32 6-4 Binomial Probability Formula

In a binomial experiment, the probability of

exactly X successes in n trials is

n!
P( X ) = p Xq n − X
(n − X )! X !
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6-33 6-4 Binomial Probability - Example

 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 ≈ 0.05.
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6-34 6-4 Binomial Probability - Example

 A survey from Teenage Research

Unlimited (Northbrook, Illinois.) 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.
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6-35 6-4 Binomial Probability - Example


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.
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6-36 6-4 Binomial Probability - Example
 A report from the Secretary of Health and
Human Services 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.
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6-37 6-4 Binomial Probability - Example

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

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 6-4 Mean, Variance, Standard Deviation
6-38 for the Binomial Distribution - Example

 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.
 µ = n⋅p = (4)(1/2) = 2.

σ 2 = n⋅p⋅q = (4)(1/2)(1/2) = 1.
 σ = 1 = 1.

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