© Boardworks Ltd 20051 of 39 © Boardworks Ltd 2005 1 of 39 AS-Level Maths: Statistics 1 for...

© Boardworks Ltd 20051 of 39 © Boardworks Ltd 20051 of 39

AS-Level Maths: Statistics 1for Edexcel

S1.5 Discrete random variables

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Discrete random variables

Introduction to discrete random variables

Cumulative distribution functions

Expectation

Variance and standard deviation for random variables

Discrete uniform distribution

Expectation algebra – some key results


The following are all examples of random variables:


The first three examples above are all discrete random variables – they all take whole number values.

In general, a random variable (r.v.) is a quantity whose value cannot be predicted with certainty before an experiment or enquiry is undertaken.

the number of heads obtained when a coin is tossed four times;

the number of prizes I win if I buy 10 tickets in a raffle;

the number of cars that pass a checkpoint in a minute;

the time (in seconds) it takes to run a 100m race.


A dice is thrown. Let X be the score obtained.

The possible outcomes of this experiment are the values 1, 2, 3, 4, 5 and 6.

These outcomes can be shown in a table, along with their corresponding probabilities:

x 1 2 3 4 5 6

P(X = x)


1

6

1

6

1

6

1

6

This table is called the probability distribution of X.

Notice that a lower case x is used to denote a particular possible outcome.

A random variable is usually denoted by a capital letter.

1

6

1

6


The probability distribution of a general discrete random variable, X, is a list or table of all its possible values, together with the corresponding probabilities:


x x1 x2 x3 … xn

P(X = x) p1 p2 p3 … pn

An important property of discrete random variables is:

i.e.

...1 2 1np p p

1ii

p

Note: the probabilities may sometimes be given as a formula rather than listed in a table.


Example: The probability distribution of a discrete random variable Y is given below:


y 0 1 2 3 4

P(Y = y) c 2c 3c 2c c

Find c and P(Y > 2).

As , c + 2c + 3c + 2c + c = 1

So, 9c = 1

Therefore c =

1ii

p

P(Y > 2) = P(Y = 3 or 4) = 2c + c = 2

9

11

9 3

19


Examination-style question: A bag contains 5 green counters and 2 red counters. John randomly takes counters from the bag, without replacement, until he draws a green counter.

The total number of counters picked out until the first green counter appears is denoted X.

Find the probability distribution of X.



G

G

G

R

R 1

5

7


2

7

5

6

1

6

P(X = 1) =

P(X = 2) =

P(X = 3) =

5

7

2 5 5

7 6 21

2 1 11

7 6 21

x 1 2 3

P(X = x)

So the probability distribution of X is:

5

7

1

21

5

21


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Expectation






The cumulative distribution function (c.d.f.), F(x) for a discrete random variable X is defined as:

x 5 10 15 20 25 30

P(X = x) 0.1 0.2 0.2 0.3 0.1 0.1


x 5 10 15 20 25 30

F(x) 0.1 0.3 0.5 0.8 0.9 1

F(x) = P(X ≤ x)F(x) = P(X ≤ x)

then it has the cumulative distribution function shown in the table below:

If X has the following probability distribution:



x 1 2 3 4

F(x) k 4k 9k 16k

Examination-style question: The cumulative distribution function, F(x), for a discrete random variable X is defined by the formula

F(x) = kx2 for x = 1, 2, 3, 4.

a) Find the value of k.

b) Find P(X > 2).

a) As x only takes the values 1, …, 4, then F(4) = 1.

So, k =

b) P(X > 2) = P(X = 3, 4) = F(4) – F(2) =

1

161

31

4 4

The c.d.f. can be shown in a table:


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Expectation




Expectation


Gambling games


The theoretical mean, μ, of a discrete random variable X is found by multiplying each possible value of X by its probability, and then adding these products together:

( )Pi i i ii i

x X x x p

Example: The probability distribution of a random variable X is:

Expectation

x –2 –1 0 1

P(X = x) 0.2 0.2 0.2 0.4

E[X] = (–2 × 0.2) + (–1 × 0.2) + (0 × 0.2) + (1 × 0.4) = –0.2

μ is also called the expectation (or expected value) of X, written E[X].


If a probability distribution is symmetrical, it is possible to write down the expectation without any calculation.

Example: The probability distribution of a random variable M is:

Expectation

m 0 10 20 30 40

P(M = m) 0.05 0.25 0.4 0.25 0.05

This distribution is symmetrical about the value M = 20.

Therefore E[M] = 20.


Examination-style question: A random variable X has the following probability distribution:

Expectation

x 0 1 2 3 4

P(X = x) 0.05 a 2a b 0.05

a) If E[X] = 1.9, find a and b.

b) Find P(0 ≤ X < 3).


0.05 + a + 2a + b + 0.05 = 1 3a + b = 0.9 (1)

E[X] = 1.9 (0×0.05) + (1a) + (2×2a) + (3b) + (4×0.05) = 1.9 5a + 3b = 1.7 (2)

Solving equations (1) and (2) simultaneously gives:a = 0.25, b = 0.15

b) P(0 ≤ X < 3) = P( X = 0, 1, 2) = 0.05 + 0.25 + 0.5 = 0.8.

Expectation

1ii

p

a) x 0 1 2 3 4

P(X = x) 0.05 a 2a b 0.05


If X is a discrete random variable, then the expectation of a function of X, g(X), is:

[ ( )] ( )E i ii

g X g x p

Example: Find E[X2 + 2X] for the following probability distribution:

Expectation of a function of X

x 1 2 3 4

x2 + 2x 3 8 15 24

P(X = x) 0.2 0.2 0.2 0.4

E[X2 + 2X] = (3 × 0.2) + (8 × 0.2) + (15 × 0.2) + (24 × 0.4)

= 0.6 + 1.6 + 3 + 9.6

= 14.8


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Expectation




Variance and standard deviation


If X is a discrete random variable, then the variance of X is given by the formula:

where μ = E[X].

Note: There is an alternative formula for Var[X] that is usually simpler to use:

i.e., the variance is the mean of the squares minus the square of the mean.


[ ] ( )2Var E[ ]X X

[ ] 2 2Var E[ ]X X

The standard deviation, σ, is the square root of Var[X].


Example: X is the random variable representing the score obtained when a normal dice is thrown. Find the mean and the standard deviation of X.

x 1 2 3 4 5 6

P(X = x) 1

6

This probability distribution is symmetrical.

Therefore, E[X] = 3.5


1

6

1

6

We can write down the probability distribution of X:

1

6

1

6

1

6


To find the variance and standard deviation, we first need E[X2]:

x2 1 4 9 16 25 36

P(X = x) 1

6

So, [ ] ...2 1 1 1 16 6 6 6

16

E 1 4 9 36

15

X

[ ]21 1

6 21112Var 15 3 2X Therefore,

So, (to 3 s.f.)[ .]Var 1 71X


1

6

1

6

1

6

1

6

1

6


Example 2: Peter drives to work every morning, but is frequently late arriving. X is the number of mornings that he is late to work in a given week.

His boss estimates that the probability distribution of X is:

x 0 1 2 3 4 5

P(X = x) 0.12 0.26 0.33 0.19 0.07 0.03

a) Find the mean number of days he is late to work per week.b) Find the standard deviation.c) His boss decides to fine him £25 if he is late once or twice

in a week and £40 if he is late more than twice in a week. Find his expected weekly fine.

d) Find the probability that he is late exactly once in a two


week period.


a)


[ ] ( . ) ( . ) ... ( . .)E 0 0 12 1 0 26 5 0 03 1 92X

[ ] ( . ) ( . ) ... ( . ) .2 2 2 2E 0 0 12 1 0 26 5 0 03 5 16X

[ ] . . .2Var 5 16 1 92 1 4736X

b)

. .1 21 1 4736 (3 s.f.)

Therefore,

So,

x 0 1 2 3 4 5

P(X = x) 0.12 0.26 0.33 0.19 0.07 0.03


y 0 25 40

P(Y = y) 0.12 0.59 0.29

c) Let Y be the random variable representing his weekly fine.


[ ] ( . ) ( . ) ( . ) .E 0 0 12 25 0 59 40 0 29 26 35Y

So, his expected weekly fine is £26.35.

d) P(late exactly once in a 2 week period) = ( . . ) ( . .. )0 12 0 26 0 2 0 06212 46 0

Not late at all in

first week

Late once in second

week

Late once in first week

Not late at all in second

week


Examination-style question: Janet takes part in a game at a school fete based on hoop-la. On each turn at the game she throws up to three hoops. She wins a sweet each time she throws a hoop over a peg. However, the turn is over as soon as a hoop misses a peg, or once she has thrown all three hoops.

On each throw, the probability of Janet getting the hoop over a peg is 0.4. Let X represent the number of sweets she wins on one turn at the game.


a) Show that P(X = 2) = 0.096, and find P(X = 0), P(X = 1) and P(X = 3).

b) Find the mean and standard deviation of X.


a) P(X = 2) = P(succeeds on first two goes, but misses on 3rd)

= 0.4 × 0.4 × 0.6 = 0.096

P(X = 0) = P(misses on 1st go) = 0.6

P(X = 1) = P(succeeds on 1st go, but then misses)

= 0.4 × 0.6 = 0.24

P(X = 3) = P(succeeds on all 3 tries) = 0.43 = 0.064


x 0 1 2 3

P(X = x) 0.6 0.24 0.096 0.064

[ ] ( . ) ( . ) ( . ) (

.

. )E 0 0 6 1 0 24 2 0 096 3 0 064

0 624

X

b) The probability distribution of X is as follows:



x 0 1 2 3

P(X = x) 0.6 0.24 0.096 0.064

[ ] ( . ) ( . ) ( . ) ( . )

.

2 2 2 2 2E 0 0 6 1 0 24 2 0 096 3 0 064

1 2

X

Using Var(X) = E[X2] – μ2, we get:

[ ] . . 2Var 1 2 0 624 0.810624X

So, σ = 0.900 (to 3 s.f.)


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Expectation






A discrete uniform distribution arises when a random variable can take a finite number of equally likely possible values. So, if X has a discrete uniform distribution taking the values x1, …, xn, its probability distribution will be:

x x1 x2 … xn

P(X = x) …

X is the score obtained when a fair six-sided dice is thrown;

X is the number on the first ball generated by the National Lottery machine;

X is the value of the digit generated by a computer’s

1n

1n

1n


Examples of random variables that have discrete uniform distributions are:

random digit generating program.


An important special case is that of a discrete uniform distribution taking the values 1, 2, …, n:

This uniform distribution is sometimes denoted by U(1, n).

The mean and variance of this particular distribution are:

E[X] = and Var[X] =

x 1 2 … n

P(X = x) …

1

2

n 2 1

12

n


1n

1n

1n



.4 1

2 52

24 1 5

12 4

X therefore has a uniform distribution, X ~ U(1, 4).

Example: A tetrahedral dice has its faces numbered 1, 2, 3 and 4. X is the score obtained when the dice is rolled.

and Var[X] =

So, E[X] =


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Expectation




Expectation algebra


There are two key results that relate to the expectation and variance of a linear function of a random variable, X:

Result 1: E[aX + b] = aE[X] + b

Result 2: Var[aX + b] = a²Var[X]

Proof of result 1:

Expectation algebra

[ ] ( )

]

( )

[E

E i i i i ii i

i i i i i ii i i i

aX b ax b p ax p bp

ax p bp

a X b

a x p b p


Proof of result 2:

Expectation algebra

[ ] ( [ ])

( )

( ) (

[

)

]2

2

2

2 2 2

Var

Var E E

E

E E

aX b aX b aX b

aX b a b

aX a a X

a X

where μ = E[X]

Result 2: Var[aX + b] = a²Var[X]

[ ] ( )2Var E[ ]X X Remember:


Example: A random variable Y has mean 5 and variance 16.

Find the mean and the variance of:

Expectation algebra

a) E[2Y] = 2E[Y] = 2 × 5 = 10

Var[2Y] = 4Var[Y] = 4 × 16 = 64

b) E[3Y – 8] = 3E[Y] – 8 = 3 × 5 – 8 = 7

Var[3Y – 8] = 9Var[Y] = 9 × 16 = 144

a) 2Y

b) 3Y – 8


Examination-style question: A random variable X has probability distribution given by:

P(X = x) = for x = 1, 2, 3

a) The probability distribution for X can be shown in a table:

x 1 2 3

P(X = x) c

The sum of the probabilities must be 1, so

Expectation algebra

c

x

2

c3

c

561 1c

6

11

a) Find c.

b) Find Var[X].

c) Find Var[6X + 2].

i.e. c =


b) The probability distribution for X therefore, is:

Expectation algebra

[ ] 6 3 18211 11 11 11E 1 2 3X

[ ]2 2 2 26 3 36211 11 11 11E 1 2 3X

] .[236 18

11 11Var 0 595X

Using

and

we get (to 3 s.f.)

c) Var[6X + 2] = 36 × 0.595 = 21.4 (to 3 s.f.)

x 1 2 3

P(X = x) 3

11

6

11

2

11

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