chapter 7 · 2019. 8. 16. · probability is approximately 0.60 that on a randomly selected day the...

Chapter 7

Random Variables

Lesson 7-1, Part 1

Discrete and Continuous

Random Variables.

Random Variable

A numerical variable whose value depends on the outcome of a chance experiment is called random variable.

A random variable associates a numerical value with each outcome of a chance experiment.

We use a capital letter like X, to denote a random value.

The probability distribution of a random variable Xgives the probability associated with each possible X value .

A random variable can be discrete or continuous.


Random Variables

A discrete variable is a variable whose values are obtained by counting. Example

Number of students present

Number of heads when flipping three coins

Student’s grade level

A continuous variable is a variable whose values are obtained by measuring. Example

Height of students in class

Time it takes to get to school

Distance traveled between classes

Discrete Random Variable

Example: Let X represents the sum of two dice

Then the probability distribution of X is as follows:

2 3 4 5 6 7 8 9 10 11 12

1 2 3 4 5 6 5 4 3 2 1( )

36 36 36 36 36 36 36 36 36 36 36

X

P X

To graph the probability distribution of a discrete random

variable, construct a probability histogram.

Probability Histogram

0

0.05

0.1

0.15

0.2

2 3 4 5 6 7 8 9 10 11 12

Probability Distribution of X

Outcome

Pro

ba

bili

ty

Example – Page 396, #7.2

A couple plans to have three children. There are 8 possible arrangements

of girls and boys. For example GGB means the first two children are

girls and the child is a boy. All 8 arrangements are (approximately equally

likely.

A). Write down all 8 arrangements of the sexes of three

children. What is the probability of any one of these

arrangements?

GGG

GGB

GBB

GBG

BBB

BGG

BBG

BGB

P GGB1 1 1 1

( )2 2 2 8


GGG

GGB

GBB

GBG

BBB

BGG

BBG

BGB

P GGB1 1 1 1

( )2 2 2 8

B). Let X be the number of girls the couple has. What is

the probability that X = 2?

P X

1 3( 2) 3

8 8


GGG

GGB

GBB

GBG

BBB

BGG

BBG

BGB

C). Starting from your work in (A), find the distribution of

X. That is what values can X take, and what are the

probabilities for each value.

Value of X:

Probability:

0 1 2 3

1/ 8 3 / 8 3 / 8 1/ 8

Lesson 7-1, Part 2


Random Variables.

Continuous Random Variable

A continuous random variable X takes all values in a given interval of numbers.

The probability distribution of a continuous random variable is shown by a density curve.

The probability that X is between an interval of numbers is the area under the density curve between the endpoints.

The probability that a continuous random variable X is exactly equal to a number is zero.


Let X be a random number between 0 and 1 produced by

the idealized uniform random number generator describe

in Example 7.3 and Figure 7.5. Find the following probabilities.

A). P X(0 0.4)

0 1

1

0.4

A (0.4 0)1

0.4

B) P X(0.4 1)

A (1 0.4)1

0.6


C). P X(0.3 0.5)

0 1

1A (0.5 0.3)1

0.2

D) P X(0.3 0.5)

A (0.5 0.3)1

0.2

E) P X(0.226 0.713)

A (0.713 0.226)1

0.487

F) What important fact about continuous random variables

does comparing your answer to (c) and (d).


F) What important fact about continuous random variables

does comparing your answer to (c) and (d).

Continuous distribution assign probabilities 0 to every

individual outcome. In this case, the probabilities in (c) and

(d) are the same because the events differ by 2 individual

values, 0.3 and 0.5, each of which has a probability 0.

Example – Uniform Density Curve

A certain probability density function is made up of two straight line

Segments. The first segment begins at the origin and goes to point

(1,1). The second segment goes from (1,1) to the point (1.50, 1).

A). Sketch the distribution function,

and verify that it is a legitimate

density curve.

1TA A A


B). Find (0 0.5)P X

C). Find ( 1)P X

(0.5)(0.50)

2 2

bhA

0

0.125


D). Find (0 1.25)P X

E). Is X an example of discrete random variable or a continuous

random variable.

(1)(1)0.50

2 2

bhA

(1.25 1)1 0.25A bh

0.50 0.250.75

TA A A

0.75

Continuous

Normal Distributions

Normal distributions are

probability distributions

N(μ, σ) is a normal

distribution with mean μ and

standard deviation σ

If X has the N(μ, σ)

distribution then

xz

μ

σ


An SRS of 400 American adult is asked, “What do you think the most

serious problem facing our schools?” Suppose that in fact 40% of all

adults would answer “violence” if asked this question. The proportion

of the sample who answered “violence” will vary in repeated sampling. In

fact, we can assign probabilities to values of using the normal density

curve with mean 0.4 and standard deviation 0.023. Use the density curve

to find the probabilities of the following events.

p̂

p̂

A). At least 45% of the sample believes that violence is the schools

most serious problem.

ˆ ˆlet violence in schools, where 0.45, 0.40, 0.023p p


A). At least 45% of the sample believes that violence is the schools

most serious problem.

P p̂( 0.45)

0.4 0.45 p̂

xz

μ 0.45 0.42.17

σ 0.023

0 2.17 Z

P z( 2.17) .0150

normalcdf E(2.17, 99,0,1)

normalcdf E(0.45, 99,0.40,0.023)

0.0149


B). Less than 35% of the sample believes that violence is the most

serious problem.

P p̂( 0.35)

0.40.35 p̂

xz

μ 0.35 0.42.17

σ 0.023

0-2.17 Z

P z( 2.17) .0150

normalcdf E( 99, 2.17,0,1)

normalcdf E( 99,0.35,0.40,0.023)

0.0149


C). The sample proportion is between 0.35 and 0.45

P p̂(0.35 0.45)

0.40.35 p̂

0-2.17 Z

P z( 2.17 2.17) .9700

normalcdf ( 2.17,2.17,0,1)

normalcdf (0.35,0.45,0.40,0.023)

0.9700

0.45

2.17

Example 1

Some areas of California are particularly earth-quake-prone.

Suppose that in one such area, 20% of all homeowners are

insured against earthquake damage. Four homeowners are

selected at random; let X denote the number among the four

who have earthquake insurance.

a) Find the distribution of X. (Hint: Let S denotes a

homeowner who has insurance and F one who dose not.

Then one possible outcome is SFSS, with probability

(0.2)(0.8)(0.2)(0.2) and associated x value of 3. There are

15 other outcomes.)

Example 1

Outcomes and Probabilities

Outcome Probability X Value Outcome Probability X Value

SSSS 0.0016 4 SFFS 0.0256 2

FSSS 0.0064 3 SFSF 0.0256 2

SFSS 0.0064 3 SSFF 0.0256 2

SSFS 0.0064 3 SFFF 0.1024 1

SSSF 0.0064 3 FSFF 0.1024 1

FFSS 0.0256 2 FFSF 0.1024 1

FSFS 0.0256 2 FFFS 0.1024 1

FSSF 0.0256 2 FFFF 0.4096 0

Example 1

Probability Distribution of X

(Homeowners with Earthquake Insurance)

X Value 0 1 2 3 4

P(X) Probability Value 0.4096 0.4096 0.1536 0.0256 0.0016

b) What is the most likely value of x?

0 and 1

c) What is the probability that at least two of the four selected

homeowners have earthquake insurance?

( 2) (2) (3) (4) 0.1536 0.0256 0.0016 0.1808P X P P P

Example 2

Let the random variable X represent the profit made on a

randomly selected day by a certain store. Assume that X is

normal with mean $460 and standard deviation $75. What

is the value of P(X > $525)

Let x = store profit, where x = $525, = $460, = $75

525

( 525) 0.1931

(525, 99,460,75)

P x

normalcdf E

460

Example 3

Let the random variable X represent the profit made on

randomly selected day by a certain store. Assume X is normal

with a mean of $427 and standard deviation $35. The

probability is approximately 0.60 that on a randomly selected

day the store will make less than Xo amount of profit. Find Xo.

Let x = store profit, where x = _____, = $427, = $35

427 Xo

xo = 435.87

invnorm(0.60, 427, 35)

We have a 60% chance of

selecting a day when the store

profit will be less than $438.87

Lesson 7.2, Part 1

Means and Variance of Random Variable

Mean of a Random Variable

Is the long-run average outcome of a

experiment.

As the number of trials of the experiment

increases, the average result of the experiment

gets closer to the mean of the random variable.

Use the notation μX for the mean.

The mean of a random variable is also called

the expected mean of X.

Mean of Discrete Random Variable

The mean of a discrete random variable X, is its weighted

average. Each value of X is weighted by its probability.

Value of X:

Probability:

k

k

x x x x

p p p p

1 2 3

1 2 3

....

....

To find the mean of X, multiply each possible value by

its probability, then add all the products:

X k k

i i

x p x p x p

E X x p

1 1 2 2μ ....

( )


Example 7.1 gives the distribution of grades (A = 4, B = 3, and so on)

in a large class as

Grade : 0 1 2 3 4

0.10 0.15 0.30 0.30 0.15Prob:

Find the average (that is, the mean) grade in this course.

Xμ 0(0.10) 1(0.15) 2(0.30) 3(0.30) 4(0.15) 2.25

Variance of a Discrete

Random Variable

If X is a discrete random variable with mean μ, then

the variance of X is

Value of X:

Probability:

k

k

x x x x

p p p p

1 2 3

1 2 3

....

....

X X X k X k

i x i

x p x p x p

x u p

2 2 22

1 1 2 2

2

σ μ μ .... μ

The standard deviation (σX) is the square root of the variance


Example 7.1 gives the distribution of grades (A = 4, B = 3, and so on)

in a large class as

Grade : 0 1 2 3 4

0.10 0.15 0.30 0.30 0.15Prob:

Find the standard deviation σX of the distribution of grades

in Exercise 7.22.

X

2 2 22

2 2

σ 0 2.25 0.10 1 2.25 0.15 2 2.25 0.30

3 2.25 0.30 4 2.25 0.15 1.3875


X

2 2 22

2 2

σ 0 2.25 0.10 1 2.25 0.15 2 2.25 0.30

3 2.25 0.30 4 2.25 0.15 1.3875

Xσ 1.3875 1.178


The Tri-State Pick 3 lottery game offers a choice of several bets. You

choose a three-digit number. The lottery commission announces

the winning three-digit number, chosen at random, at the end of each

day. The “box” pays $83.33 if the number you choose has the same

digits as the winning number, in any order. Find the expected payoff

for $1 bet on the box. (Assume that you chose a number having

three different digits.)

If my number is abc, then of the (10)3 = 1000 three-digit

numbers, there are six – abc, acb, bac, bca, cab, cba – for

Which you will win the box.

P W6

( ) 0.0061000

P L( ) 1 0.006 0.994


The Tri-State Pick 3 lottery game offers a choice of several bets. You

choose a three-digit number. The lottery commission announces

the winning three-digit number, chosen at random, at the end of each

day. The “box” pays $83.33 if the number you choose has the same

digits as the winning number, in any order. Find the expected payoff

for $1 bet on the box. (Assume that you chose a number having

three different digits.)

P W6

( ) 0.0061000

P L( ) 1 0.006 0.994

Payoff X $0 $83.33

Probability 0.994 0.006


Payoff X $0 $83.33

Probability 0.994 0.006

Find the expected payoff on a $1 bet

Xμ 0 0.994 83.33 0.006 $0.50

The casino keeps 50 cents from each dollar bet in the long

run, since the expected payoff is 50 cents.

Payoff X $ – 1 $83.33

Probability 0.994 0.006 Xμ 0.50

Law of Large Numbers

As the number of observations

increases, the mean of the observed

values of, , approaches the mean of

the population, μ.

The more variation in the outcomes,

the more trials are need to ensure

that is close to μ.

x

x


A). A gambler knows that red and black are equally likely to occur on

each pin of a roulette wheel. He observes five consecutive reds

and bets heavily on red at the next spin. Asked why, he says that

“red is hot” and that the run of reds is likely to continue. Explain

to the gambler what is wrong with this reasoning.

The wheel is not affected by it past outcomes – it has

no memory; outcomes are independent. So on any

one spin, black and red remain equally likely


B). After hearing you explain why red and black remain equally probable

after five reds on the roulette wheel, the gambler moves to poker game.

He is dealt five straight red cards. he remembers what you said

and assume that the next card dealt in the same hand it equally likely

to red or black. Is the gambler right or wrong? Why?

Removing a card changes the composition of the

remaining deck, so successive draws are not independent.

If you hold 5 red cards, the deck now contains

5 fewer reds cards, so your chance of another red

decreases.

Lesson 7.2, Part 2

Rules for Means and Variance

Rules for Means

a bX Xa bμ μ

If X is a random variable and a and b are

fixed numbers, then

If X and Y are random variables, then

X Y X Yμ μ μ

Rules for Variances

a bX Xb2 2 2σ σ

If X is a random variable and a and b are

fixed numbers, then

If X and Y are independent random variables, then

X Y X Y

X Y X Y

2 2 2

2 2 2

σ σ σ

σ σ σ

Example 1 – Page Rules for

Means and Variance

Given independent random variables with mean and standard deviations

are shown, find the means and standard deviations of each of these

variables.Mean SD

X 120 12

Y 300 16A) 0.8Y

0.8

2 22

0.8

0.8

0.8(300) 240

0.8 16 163.84

163.84 12.80

Y

Y

Y

B) 2X – 100

2 100

2 22

2 100

2 100

2(120) 100 140

2 12 576

576 24

X

X

X

Example 1 – Page Rules for

Means and Variance

Given independent random variables with mean and standard deviations

are shown, find the means and standard deviations of each of these

variables.Mean SD

X 120 12

Y 300 16C) X + 2Y

D) 3X – Y

2

2 2 22

2

2

120 2(300) 720

12 2 16 1168

1168 34.18

X Y

X Y

X Y

3

2 2 22

3

3

3(120) 300 60

3 12 16 1552

1552 39.40

X Y

X Y

X Y

Example 1: Rules for Means

Suppose the equation Y = 20 + 10X converts a PSAT math

score, X, into an SAT math score, Y. Suppose the average

PSAT math score is 48. What is the average SAT math

score?

48X

a bx Xa b

20 100 20 10(48)

500

X

Example 2: Rules for Means

Let μX = 625 represents the average SAT math score.

X Y X Y

Let μy = 590 represents the average SAT verbal score.

represents the average combined

SAT score. Then 625 590 1215X Y X Y

is the average combined total SAT score.

Example 3

Rules for Variances

Suppose the equation Y = 20 + 10X converts a PSAT math

score, X, into an SAT math score, Y. Suppose the standard

deviation for the PSAT math score is 1.5 points. What is the

standard deviation for the SAT math score?

2 2

2 2 2

(1.5) 2.25X

a bX Xb

2 2

20 100 (10) (2.25)

225

15

X

X


Laboratory data show that the time required to complete two chemical

reactions in a production process varies. The first reaction has a mean

time of 40 minutes and a standard deviation of 2 minutes: the second

has a mean time of 25 minutes and a standard deviation of 1 minute. The

reactions are run in sequence during production. There is fixed period

of 5 minutes between them as the product of the first reaction is pumped

into the vessel where the second reaction will take place. What is the

mean time required for the entire process?

X X1 25 40 5 25 70 minutesTotal Mean


Use Examples 7.7 (Page 411) and 7.10 (Page 419) concerning a

probabilistic projection of sales and profits by an electronic firm,

Gain Communications.

A). Find the variance and standard deviation of the estimated sales Y

of Gain’s civilian unit, using the distribution and mean from Example

7.10

Yμ 445 unitsY = Units sold 300 500 750

Probability 0.4 0.5 0.1

Y

Y

2 2 22σ 300 445 0.4 500 445 0.5 750 445 0.1 19225

σ 19225 138.65 units


B). Because the military budget and the civilian economy are not closely

linked, Gain is willing to assume that its military and civilian sales

vary independently. Combine you results from (a) with the results

for the military unit from Example 7.10 to obtain the standard deviation

of total sales X + Y

Y

Y

μ 445

σ 138.65

units X = Units sold 1000 3000 5000 10000

Probability 0.1 0.3 0.4 0.2

X

X

2 2 22

2

σ 1000 5000 0.1 3000 5000 0.3 5000 5000 0.4

10000 5000 0.2 7,800,000

σ 7800000 2792.85

units

units

Xμ 5000 units


B). Because the military budget and the civilian economy are not closely

linked, Gain is willing to assume that its military and civilian sales

vary independently. Combine you results from (a) with the results

for the military unit from Example 7.10 to obtain the standard deviation

of total sales X + Y

Y

Y

μ 445

σ 138.65

units

unitsunits

Xμ 5000 units

Xσ 2792.85

X Y X Y

2 2 2 2 2σ σ σ 138.65 2792.85 7,819,225

σ 7819234.95 2796.29


C). Find the standard deviation of the estimated profit, Z = 2000X + 3500Y

Y

Y

μ 445

σ 138.65

units

unitsunitsXμ 5000 units

Xσ 2792.85

X Y X Y X Yb b2 2 2 2 2 2 2

2000 3500 2000 3500

2 2 2 2

13

σ σ σ σ σ

(2000) 2792.85 (3500) 138.65

3.1434 10

13σ 3.144 10 $5,606,738

chapter 7 · 2019. 8. 16. · probability is approximately 0.60 that on a randomly selected day the...

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