Chapter 3

Discrete Random Variables and Probability Distributions

Chapter 3A

Variables that are random; what will they

think of next?

The Road Ahead

today

A Random Variable (RV)

Variable whose observed value is determined by chance

Variable that takes on values in accordance with some probability distribution

Discrete Random Variables have a finite or countably infinite range

Numerical outcome from a random experiment

A mapping from a sample space to a subset of the real numbers

Some Discrete Random Variables

The number of nonconforming solder connections on a printed circuit board.

In a voice communication system with 50 lines, the number of lines in use at a particular time.

A batch of 500 machined parts contains 10 that do not conform to customer requirements. Parts are selected successively, without replacement, until a nonconforming part is obtained. The RV is the number of parts selected.

The RV is the number of demands in a month for a product in inventory.

The RV is the number of customer arrivals per hour at a local bank.

The number of accidents per week observed in a factory.

The Mapping Illustrated – toss a pair of dice

Let X = a random variable, the sum resulting from the toss of two fair dice; X = 2, 3, …, 12

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

S =

The Mapping Illustrated – toss a pair of dice

   

2 (1,1) 1

3 (1,2) , (2,1) 2

4 (1,3) , (2,2) , (3,1) 3

5 (1,4) , (2,3) , (3,2) , (4,1) 4

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

7 (1,6) , (2,5) , (3,4) , (4,3) , (5,2) , (6,1) 6

8 (2,6) , (3,5) , (4,4) , (5,3) , (6,2) 5

9 (3,6) , (4,5) , (5,4) , (6,3) 4

10 (4,6) , (5,5) , (6,4) 3

11 (5,6) , (6,5) 2

12 (6,6) 1

Total   36

X = RV, the outcome from rolling a pair of dice number of ways

sample space

The Mapping Illustrated – toss a pair of dice

Let X = a random variable, the sum resulting from the toss of two fair dice; X = 2, 3, …, 12

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

x 2 3 4 5 6 7 8 9 10 11 12p(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

= S

Pr{X = x} = f(x), f(x) is called the Probability Mass Function (PMF)

f(x)

Probability Histogram for the Random Variable X

Probability Mass Function

Example #1 Let x = a discrete random variable, the number of

accidents per week in the Axe E. Dentt manufacturing plant.

Given: 1( ) Pr{ } ; 0,1,2,3,4

15

xf x X x x

Pr(X = 0) = f(0) = 1/15Pr(X = 1) = f(1) = 2/15Pr(X = 2) = f(2) = 3/15Pr(X = 3) = f(3) = 4/15Pr(X = 4) = f(4) = 5/15

4

0

( ) 0

( ) 1x

f x

f x

Example #2

Let x = a discrete random variable, the number of days to receive a package from a Website distributor when requesting expedited delivery.

.1 if 2days

.4 if 3days

( ) .3 if 4days

.1 if 5days

.1 if 6days

x

x

f x x

x

x

Example #3 Let x = a discrete random variable, the number of

units produced before a reject occurs. The probability of a reject occurring is 1/5.

11 4

, 1,2,3...( ) 5 5

0 otherwise

x

xf x

1

1 0

1 4 1 4 1 11

45 5 5 5 5 15

x x

x x

note:

Find: Pr{X 10}Pr{20 X 30}Pr{X 15}

geometric series

3-3 Cumulative Distribution Function (CDF)

Definition

Example #3 revisited

Find: Pr{X 10} = F(10) = .892626 Pr{20 X 30} = F(30) – F(20) = (1-.830) – (1-.820) = .99876 - .98847 = .0103Pr{X 15} = 1 – F(14) = 1 – (1 - .814) = .814 = .04398

1 1

1 1 0

1 4 1 4( ) ( )

5 5 5 5

41

1 451

45 515

i ix x x

i i i

x

x

F x f i

1

1

n

n

a rS

r

x f(x) F(x)

1 0.2 0.22 0.16 0.363 0.128 0.4884 0.1024 0.59045 0.08192 0.672326 0.065536 0.7378567 0.052429 0.7902858 0.041943 0.8322289 0.033554 0.86578210 0.026844 0.89262611 0.021475 0.91410112 0.01718 0.93128113 0.013744 0.94502414 0.010995 0.9560215 0.008796 0.96481616 0.007037 0.97185317 0.005629 0.97748218 0.004504 0.98198619 0.003603 0.98558820 0.002882 0.988471

Problem 3-28

Determine the cumulative distribution function of the following R.V.’s [P(xi) = 1/6 for all xi]:

Outcome a b c d e f x 0 0 1.5 1.5 2 3

0, 0

1/ 3, 0 1.5

( ) 2 / 3, 1.5 2

5/ 6, 2 <3

1, 3

x

x

F x x

x

x

F(x)1.0

2/3

1/3

0 1.5 2 3

Don’t confuse discrete with

integer!

Problem 3-35

Verify the following function is a CDF & determine the PMF & requested probabilities:

P(X < 3) = P(1 < X < 2) = P(X < 2) = P(X > 2) =

x

x

x

xF

31

315.

10

)(

PMF:f(0) = F(0) = 0f(1) = F(1) – F(0) = .5f(2) = F(2) – F(1) = .5 - .5 = 0f(3) = F(3) – F(2) = 1 – .5 = .5

1

.5

.5

.5

Example 3-8

Example 3-8

Figure 3-4 Cumulative distribution function for Example 3-8.

A Fish Tale –A Transition to the next concept

These probability distributions are great. But what if my boss wants to know how many fish I

will sell today? I need one number not a entire distribution.

Let Z = a discrete random variable, thenumber of fish sold in one day.

.1 if 0

.3 if 1( )

.5 if 2

.1 if 3

z

zf z

z

z

The Mean Number of Fish

Let Z = a discrete random variable, thenumber of fish sold in one day.

.1 if 0

.3 if 1( )

.5 if 2

.1 if 3

x

xf z

x

x

( ) (.1)(0) (.3)(1) (.5)(2) (.1)(3) 1.6E Z

Expect to sell 1.6 fish on the average!

3-4 Mean and Variance of a Discrete Random Variable

Definition

Mean of a discrete R.V.

The mean of a discrete R.V. uses the probability of each discrete observation to weight that observation:

1 1 2 2( ) ( ) .... ( )n nP X x x P X x x P X x x

( ) ( )x

E X x f x

Variance of a discrete R.V.

The variance of a discrete R.V. X also uses probability to weight each observation.

2 2 2 21 1 2 2( )( ) ( )( ) .... ( )( )n nP X x x P X x x P X x x

2( ) ( ) ( )x

V X x f x 2 2( )

x

x f x

( )V X

We don’t do many derivations, but it needs to be clear to you how we get from line two to line three. Try it!

Problem 3-40

Determine the mean and variance P(xi=1/6) for all i:

Outcome a b c d e f x 0 0 1.5 1.5 2 3

= 0(1/3) + 1.5(1/3) + 2(1/6) + 3(1/6) = 4/3

2 2 2( ) ( )x

V X x f x

x

xxfXE )()(

= 02(1/3) + 1.52(1/3) + 22(1/6) + 32(1/6) – (4/3)2

= 1.139

First Two Moments do not Determine Distribution

The mean is the first moment. The variance is the second central moment –

second moment about the mean. Two entirely different distributions can have

identical mean and variance. Very often though the first two moments

give sufficient information to do effective modeling.

Yet Another ProblemDetermine the mean & variance of:

2 1( ) , 0,1,2,3,4

25

xf x x

( ) ( )

x

E X x f x

2*0 1 2*1 1 2*2 1 2*3 1 2*4 10 1 2 3 4

25 25 25 25 25

3 10 21 36

0 2.825 25 25 25

2 2 2( ) ( )x

V X x f x 2 2 2 2 2 21 3 5 7 9

0 1 2 3 4 (2.8) 1.3625 25 25 25 25

3-4 Mean and Variance of a Discrete Random Variable

Figure 3-5 A probability distribution can be viewed as a loading with the mean equal to the balance point. Parts (a) and (b) illustrate equal means, but Part (a) illustrates a larger variance.

3-4 Mean and Variance of a Discrete Random Variable

Figure 3-6 The probability distribution illustrated in Parts (a) and (b) differ even though they have equal means and equal variances.

Example 3-11

3-4 Mean and Variance of a Discrete Random Variable

Expected Value of a Function of a Discrete Random Variable

))(())(( xEhxhE Unless the function is linear

A Derivation

; [ ]

[ ] [ ] ( )

( ) ( ) ( ) ( )

[ ]

x

x

x x x

y x

Y a bX E X

E Y E a bX a bx f x

af x bxf x a f x b xf x

a bE X a b

More on Expected Values It costs Axe E. Dent $700 a week to maintain a safety

officer and another $340 for each accident that the safety officer must process. What is the expected weekly cost?

Let x = a discrete random variable, the number of accidents per week in the Axe E. Dentt manufacturing plant.

Let Y = a discrete random variable, the weekly cost of maintaining a safety officer.

4

0

1( ) Pr{ } ; 0,1,2,3,4

151 8

[ ]15 3

8[ ] 700 340 $1,606.67

3

xx

y

xf x X x x

xE X x

E Y

Y = 700 + 340X E[Y] = 700 + 340 E[X]

But Look Here

; [ ]

[ ] [ ] ( )

1 1( ) ( )

x

x

x x x

bY E X

Xb b

E Y E f xX x

bb f x b f x

x x

Keep Looking…X f(x) 1/X1 0.25 12 0.2 0.53 0.15 0.3334 0.1 0.255 0.3 0.2

sum 1E[X] = 3 0.485 = E[1/X]

1 1 1.485

3x

EX

A most important lesson has

been learned here today.

.25(1) + .2 (.5) + .15 (.333) + .1 (.25) + .3 (.2) = .485

What about the Variance?

2

2

2

2

2 22

22 2 2

where [ ]

[ ] [ ]

x

y

y

x

x x

x x

Y a bX Var X

Var Y Var a bX

E Y

E a bX a b

E bX b E b X

b E x b

Yet another insight… Analogous to the mean’s being the center

of gravity of a distribution of mass, the variance represents, in terminology of mechanics, the moment of inertia.

The moment of inertia of an object about a given axis describes how difficult it is to change its angular motion about that axis.

Bonus Topic

Let X = a random variable, the number of points scored with first down and ten yards to go, at discrete points on the playing field.

Number of outcomes is 103 touchdown +7 field goal + 3 safety -2 opponent’s touchdown -7 turning the ball over to the opponent at any of 99

possible points on the field

More of that bonus

Based upon a study of 2,852 first-and-ten plays by Virgil Carter and Robert Machol:

Field Position

Expected point value

95 -1.24585 -0.63775 0.23665 0.92355 1.53845 2.39235 3.16725 3.68115 4.5725 6.041

Next Class

Theoretical Discrete Distributions Uniform Binomial Geometric Poisson

and so much more…

ENM 500 studentshurrying to class.