Discrete Random Variables © Christine Crisp “Teach A Level Maths” Statistics 1

Discrete Random Discrete Random VariablesVariables

© Christine Crisp

““Teach A Level Teach A Level Maths”Maths”

Statistics 1Statistics 1

Discrete Random Variables

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Statistics 1

AQA

EDEXCEL

OCR


Suppose we roll an ordinary 6-sided die sixty times and record the number of ones, twos, etc.We might get Number on

die1 2 3 4 5 6

Frequency 12 9 11 10 7 11If I asked you what you might expect to happen if we went on rolling the die you might say that you would expect roughly the same number of ones, twos, etc.

In saying this, you would be using a perfectly reasonable model.Models in Statistics describe situations

and are used to make predictions.


We could write the model out as a table:

6

1

6

16

1

6

1

6

1

6

1

x gives the value of the number shown on the die.

P (X = )xP

1 2 3 4 5 6x

It is a variable which can be any value from 1 to 6.We let X be a description of the

variable, so:“ X is the number shown on the face of the die”

If x = 1, for example, we get P(X = 1) which means the probability that the number shown on the die is 1.

We label the 2nd row .P (X = )x


So, we have

6

1

6

16

1

6

1

6

1

6

11 2 3 4 5 6x

P(X = )1


So, we have

6

1

6

16

1

6

1

6

1

6

11 2 3 4 5 6x

P(X = )2


So, we have

6

1

6

16

1

6

1

6

1

6

11 2 3 4 5 6x

P(X = )3


So, we have

6

1

6

16

1

6

1

6

1

6

11 2 3 4 5 6x

P(X = )4


So, we have

6

1

6

16

1

6

1

6

1

6

11 2 3 4 5 6x

P(X = )5


So, we have

6

1

6

16

1

6

1

6

1

6

11 2 3 4 5 6x

P(X = )6


So, we have

6

1

6

16

1

6

1

6

1

6

11 2 3 4 5 6x

P(X = )x


P(X = )x

The sum from 1 to 6 of the probabilities of all the values of X, ( x = 1, 2, 3, 4, 5, 6 )

So, we have

6

1

6

16

1

6

1

6

1

6

11 2 3 4 5 6x

6

1x

This table shows the probability distribution of X.

is the Greek capital letter S and stands for Sum

So we can write

If we add up ( sum ) the probabilities we get

1

equals 1

1)( xX P


A variable where the sum of the probabilities of all its possible values is equal to 1 is called a random variable ( r.v. ).

So, in our example, X is the random variable “ the number shown on the face of the die”

1)( xXP

We can usually see what values the random variable can have, so we don’t need to show them on the summation sign.So, we often writeX is an example of a discrete random variable. It takes certain values only.In the example these values were the integers from 1 to 6. ( In exercises the numbers are often integers but they don’t have to be. )

Discrete Random VariablesSUMMARY• A statistical model uses probabilities to

describe a situation and to make predictions.

• A probability distribution gives the probabilities for a random variable.

• If X is a discrete random variable, then

1)( xXP

• A variable where the sum of the probabilities of all its possible values is equal to 1 is called a random variable (r.v.). If X takes only certain values in an interval, X is a discrete random variable.

X describes the r.v.x gives the values of the r.v.

N.B.


1)( xXPWe want to show that , so we

need to find the probabilities of getting 0, 1

or 2 sixes.

6,6,6,6,6,6,6,6 ////

Using for “not a 6 ” we can write the possibilities as

/6

e.g. 1 Let X be the variable “ the number of sixes showing when 2 dice are rolled”. Show that X is a random variable and write its probability distribution in a table.Solution:

We can have 0 sixes, 1 six or 2 sixes:

Then, )0(XP )6,6( //P 6

5

6

5

)1(XP )6,6()6,6( // PP 6

5

6

1

6

1

6

5

36

10

)2(XP )6,6(P 6

1

6

1

36

1

36

25

Tip: It will be easier to add the fractions if we don’t cancel


The probability table is

P 36

25

36

1036

10 1 2x

(X = )x

36

1

36

10

36

251

36

36

,36

25)0( XP ,

36

10)1( XP

36

1)2( XP

1)( xXPSince , X is a random variable.

So, )( xXP

5

18



P 36

25

18

5

36

10 1 2x

(X = )x

36

1

36

10

36

251

36

36

,36

25)0( XP ,

36

10)1( XP

36

1)2( XP

1)( xXPSince , X is a random variable.

So, )( xXP

This is an example of a discrete random variable because the variable takes only some values in an interval rather than every value.


But, . . . can be replaced by , . . .

the probabilities of getting 1, 2, . . .

,1

f

f1p

f

f22p1st x-value 1st

frequency

f

xfx

The Mean of a Discrete Random VariableWe can find the mean of a discrete random

variable in a similar way to that used for data. Suppose we take our first example of rolling a die.Number on

die1 2 3 4 5 6

Frequency 12 9 11 10 7 11

The mean is given by

f

fxfx ...2211

So, the mean

...2211 pxpx px


When dealing with a model, we use the letter for the mean (the greek letter m).

pxWe write

)( xXxP

or, more often, replacing p by , )( xXP

Notation for the Mean of a Discrete Random Variable

Instead of , we can also write E(X).This notation comes from the idea of the mean being the Expected value of the r.v. X.

pronounced “mew”


When dealing with a model, we use the letter for the mean (the greek letter m).

We write

)( xXxP


Instead of , we can also write E(X).

( Think of this as being what we expect to get on average ).

pronounced “mew”

This notation comes from the idea of the mean being the Expected value of the r.v. X.

px

Notation for the Mean of a Discrete Random Variable


e.g. 1. A random variable X has the probability distribution

P 4

1

2

1

1 5 10x

(X = )x p

Find (a) the value of p and (b) the mean of X.

Solution:

(a) Since X is a discrete r.v., 1)( xXP

121

41 p

41p

(b) mean, )( xXxP 411

41 1051 2 4

21

Tip: Always check that your value of the mean lies within the range of the given values of x. Here, or 5·25, does lie between 1 and 10.

421


The probabilities in a probability distribution can sometimes be given by a formula.

3,2,16

)( xx

xXP for

e.g. 1. Write out a probability distribution table for the r.v. X where

The formula is called a probability density function ( p.d.f. ).

Solution:

6

1

6

2P

1 2 3x

(X = )x 6

31

3 2

1


The probabilities in a probability distribution can sometimes be given by a formula.

3,2,16

)( xx

xXP for

e.g. 1. Write out a probability distribution table for the r.v. X where

The formula is called a probability density function ( p.d.f. ).

Solution:

6

13

1P

1 2 3x

(X = )x 2

1

These probabilities can be shown on a diagram.


1 32

3

1

6

1

2

1

x

)( xXP

6

13

1P

1 2 3x

(X = )x 2

1

This is called a stick diagram.


4,3,2,1)( xkxxXP for

e.g. 2. Find the value of the constant k for the random variable X with p.d.f. given by

Solution:Since X is a discrete random variable,

1)( xXP

So, 14321 kkkk

110 k

10 k


SUMMARY• The mean, , of a discrete random

variable is given by

)( xXxP

• The mean is also referred to as the expectation or expected value of the r.v.

• can be written as E(X)• The probabilities can be given by a formula called the probability density function ( p.d.f. )• An unknown constant in the p.d.f. can be found by using

1)( xXP


Exercise

1. The tables show the probability distributions of 2 random variables. For each, find (i) the value of p (ii) the mean value.

(a) (b)

P 6

13

1

1 2 3x

(X = )x p P 30 60

0 1 2x

(X = )x p

2. Write out the probability distribution for the random variable, X, where the probability distribution function is

4,3,2,110

)( xx

xXP for


Solution:

1(a)

P 6

1

3

1

1 2 3x

(X = )x p

(b)

P 30 60

0 1 2x

(X = )x p

13

1

6

1p

3

7

6

14

2

1p

10 p

16030 p

31

2

13

3

12

6

11

602101300 mean,

mean,

)( xXxP

X is a random variable: 1)( xXP


1 2x

P(X = x )

3 4


4,3,2,110

)( xx

xXP for

Solution:

10

1

10

2

10

3

10

41

5 5

2


1 2x

P(X = x )

3 4


4,3,2,110

)( xx

xXP for

Solution:

10

1

5

1

10

3

5

2


Exercise

3,2,1)( xx

kxXP for

3. Find the exact value of the constant k for the random variable X with p.d.f. given by

Solution:

Since X is a discrete random variable, 1)( xXP

So,

1321

kkk

16

236

kkk

16

11

k 11

6k


Variance of a Discrete Random VariableThe variance of a discrete random variable is

found in a similar way to the one we used for the mean.

22

2 xf

fxs

variance

For a frequency distribution, the formula is

222

12 ...

xf

fxfx

Replacing by etc.

gives f

f11p

22

221

21

2 ... xpxpxs


22

21

22 ... pxpx

22

221

21

2 ... xpxpxs

So,

222 )( xXPx

But we must replace by and we replace s by the letter ( which is the Greek lowercase s, pronounced sigma ).

x

The variance of X is also written as Var(X).

( Notice that this expression contains the Greek capital S, , and the lowercase s, . )


1910

100

Tip: With a bit of practice you’ll find you can simplify the fractions without a calculator. It’s quicker and more accurate. Try these before you see my answer.

Solution:

222 )( xXPx

e.g. 1 Find the variance of X for the following:

1 2x

P(X = x )

3 4

10

1

10

2

10

3

10

4

222 )( xXPx

10

44

10

33

10

22

10

11

We first need to find the mean,

310

30

)( xXxP

222222 )3(10

44

10

33

10

22

10

11

910

64

10

27

10

8

10

1


SUMMARY

222 )()(Var xXPxX

• The variance, , of a discrete random variable is given by

2

• The mean of a discrete random variable is given by

)()(E xXxPX

For probability distributions ( models ) use and ( the Greek alphabet ).

2

N.B. For frequency distributions use and for the mean and variance ( the “English” alphabet ).

x 2s

Discrete Random VariablesExercis

e1. Find the variance of X for each of the following:(a) (b)

P 6

16

2

1 2 3x

(X = )x P 30 60

0 1 2x

(X = )x6

310

Solution:

(a)

222 )( xXPx2

222

3

7

6

33

6

22

6

11

)( xXxP

6

33

6

22

6

11

6

14

7

3 3

7

9

49

6

27

6

8

6

1

9

496

9

5


Exercise

(b)

P 30 60

0 1 2x

(X = )x 10

Solution:

222 )( xXPx

222 31602101

)( xXxP

602101 31

69152 810


The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Discrete Random VariablesSUMMARY• A statistical model uses probabilities to

describe a situation and to make predictions.

• A probability distribution gives the probabilities for a model.

• If X is a discrete random variable ( r.v. ), then 1)( xXP

• A variable where the sum of the probabilities of all its possible values is equal to 1 is called a random variable (r.v.). If X takes only certain values in an interval, X is a discrete r.v.

X describes the r.v.x gives the values of the r.v.

N.B.


f

xfx

The Mean of a Discrete Random VariableWe can find the mean of a discrete random variable in a similar way to that for data. Suppose we take our first example of rolling a die.

1171011912Frequency

654321Number on die

The mean is given by

f

fxfx ...2211

So, the mean

...2211 pxpx px

But, . . . can be replaced by , . . .

the probabilities of getting 1, 2, . . .

,1

f

f1p

f

f22p


When dealing with a discrete random variable, we use the letter ( pronounced mew ) for the mean (the greek letter m).

xpWe write

)( xXxP


Notation for the Mean of a discrete Random Variable

Instead of , we can also write E(X).The notation comes from the idea of the mean being the Expected value of the r.v. X.

( Think of this as being what we expect to get on average ).


6,6,6,6,6,6,6,6 ////

Using for “ not a 6 “ we can write the possibilities as

/6

e.g. 1 Let X be the variable “ the number of sixes showing when 2 dice are rolled”. Show that X is a random variable and write its probability distribution in a table.

Solution:

We can have 0 sixes, 1 six or 2 sixes:

Then, )0(XP )6,6( //P 6

5

6

5

)1(XP )6,6()6,6( // PP 6

5

6

1

6

1

6

5

36

10

)2(XP )6,6(P 6

1

6

1

36

1

36

25



36

1

36

10

36

251

36

36

,36

25)0( XP ,

36

10)1( XP

36

1)2( XP

So, )( xXP

P 36

25

18

5

36

10 1 2x

(X = )x


SUMMARY• The mean, , of a discrete random

variable is given by

)( xXxP

• The mean is also referred to as the expectation or expected value of the r.v.

• can be written as E(X)

• The probabilities can be given by a formula called the probability density function ( p.d.f. )


Variance of a Discrete Random Variable

22

2 xf

fxs

variance

For a frequency distribution, the formula is

22

22

1 ...x

f

xfxf

Replacing by etc.

gives f

f11p

2222

211

2 ... xxpxps

222

21

2 ... xpxp222 )( xXPx

But for a random variable, we must replace by and we replace s by the letter (the greek s, pronounced sigma ). So,

x

The variance of X is also written as Var(X).


1910

100

Solution:

222 )( xXPx

e.g. 1 Find the variance of X for the following:

1 2x

P(X = x )

3 4

10

1

10

2

10

3

10

4

222 )( xXPx

10

44

10

33

10

22

10

11

We first need to find the mean,

310

30

)( xXxP

222222 )3(10

44

10

33

10

22

10

11

910

64

10

27

10

8

10

1


SUMMARY

222 )()(Var xXPxX

• The variance, , of a discrete random variable is given by

2

• The mean, , of a discrete random variable is given by

)()(E xXxPX

For probability distributions ( models ) use and ( the Greek alphabet ).

2

N.B. For frequency distributions use and for the mean and variance ( the “English” alphabet ).

x 2s

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Discrete Random Variables © Christine Crisp “Teach A Level Maths” Statistics 1