Download S1.5 Discrete random variables

AS-Level Maths:
Statistics 1
for Edexcel
S1.5 Discrete random
variables
This icon indicates the slide contains activities created in Flash. These activities are not editable.
For more detailed instructions, see the Getting Started presentation.
11 of
of 39
39
© Boardworks Ltd 2005
Discrete random variables
Contents
Introduction to discrete random variables
Cumulative distribution functions
Expectation
Variance and standard deviation for random
variables
Discrete uniform distribution
Expectation algebra – some key results
22 of
of 39
39
© Boardworks Ltd 2005
Discrete random variables
The following are all examples of random variables:
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.
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 first three examples above are all discrete random
variables – they all take whole number values.
3 of 39
© Boardworks Ltd 2005
Discrete random variables
A dice is thrown. Let X be the score obtained.
The possible outcomes
of this experiment are the values 1, 2,
A random variable
3, 4, 5 and 6. is usually denoted
by a capital letter.
These outcomes can be shown in a table, along with their
corresponding probabilities:
x
P(X = x)
1
1
6
2
1
6
3
1
6
4
1
6
5
1
6
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.
4 of 39
© Boardworks Ltd 2005
Discrete random variables
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:
p1  p2  ...  pn  1
p
i.e.
i
1
i
Note: the probabilities may sometimes be
given as a formula rather than listed in a table.
5 of 39
© Boardworks Ltd 2005
Discrete random variables
Example: The probability distribution of a discrete
random variable Y is given below:
y
P(Y = y)
0
c
1
2c
2
3c
3
2c
4
c
Find c and P(Y > 2).
As  pi  1 ,
c + 2c + 3c + 2c + c = 1
So,
9c = 1
i
Therefore
c = 91
2 1 1
P(Y > 2) = P(Y = 3 or 4) = 2c + c =  
9 9 3
6 of 39
© Boardworks Ltd 2005
Discrete random variables
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.
7 of 39
© Boardworks Ltd 2005
Discrete random variables
5
7
P(X = 1) = 5
7
G
2
7
5
6
2 5 5
P(X = 2) =  
7 6 21
G
R
2 1
1
1
1
R
G P(X = 3) =   1 
7 6
21
6
So the probability distribution of X is:
x
P(X = x)
8 of 39
1
5
7
2
5
21
3
1
21
© Boardworks Ltd 2005
Cumulative distribution functions
Contents
Introduction to discrete random variables
Cumulative distribution functions
Expectation
Variance and standard deviation for random
variables
Discrete uniform distribution
Expectation algebra – some key results
99 of
of 39
39
© Boardworks Ltd 2005
Cumulative distribution functions
The cumulative distribution function (c.d.f.), F(x) for a
discrete random variable X is defined as:
F(x) = P(X ≤ x)
If X has the following probability distribution:
x
P(X = x)
5
10
15
20
25
30
0.1
0.2
0.2
0.3
0.1
0.1
then it has the cumulative distribution function shown
in the table below:
x
F(x)
10 of 39
5
10 15 20 25
0.1 0.3 0.5 0.8 0.9
30
1
© Boardworks Ltd 2005
Cumulative distribution functions
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).
The c.d.f. can be shown in a table:
x
F(x)
1
k
a) As x only takes the values
1, …, 4, then F(4) = 1.
1
So, k = 16
1 3
b) P(X > 2) = P(X = 3, 4) = F(4) – F(2) = 1 
4 4
11 of 39
2
4k
3
4
9k 16k
© Boardworks Ltd 2005
Expectation
Contents
Introduction to discrete random variables
Cumulative distribution functions
Expectation
Variance and standard deviation for random
variables
Discrete uniform distribution
Expectation algebra – some key results
12 of 39
© Boardworks Ltd 2005
Gambling games
13 of 39
© Boardworks Ltd 2005
Gambling games
14 of 39
© Boardworks Ltd 2005
Expectation
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:
   xi  P( X  xi )   xi pi
i
i
μ is also called the expectation (or expected value) of
X, written E[X].
Example: The probability distribution of a random variable X
is:
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
15 of 39
© Boardworks Ltd 2005
Expectation
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:
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.
16 of 39
© Boardworks Ltd 2005
Expectation
Examination-style question: A random variable X has the
following probability distribution:
x
P(X = x)
0
0.05
1
a
2
2a
3
b
4
0.05
a) If E[X] = 1.9, find a and b.
b) Find P(0 ≤ X < 3).
17 of 39
© Boardworks Ltd 2005
Expectation
a)
x
P(X = x)
p
i
i
1 

0
0.05
1
a
2
2a
3
b
4
0.05
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.
18 of 39
© Boardworks Ltd 2005
Expectation of a function of X
If X is a discrete random variable, then the expectation of a
function of X, g(X), is:
E[ g ( X )]   g ( xi ) pi
i
Example: Find E[X2 + 2X] for the following probability
distribution:
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
19 of 39
© Boardworks Ltd 2005
Variance and standard deviation
Contents
Introduction to discrete random variables
Cumulative distribution functions
Expectation
Variance and standard deviation for random
variables
Discrete uniform distribution
Expectation algebra – some key results
20 of 39
© Boardworks Ltd 2005
Variance and standard deviation
If X is a discrete random variable, then the variance of X is
given by the formula:
Var[ X ]  E[( X   )2 ]
where μ = E[X].
The standard deviation, σ, is the square root of Var[X].
Note: There is an alternative formula for Var[X]
that is usually simpler to use:
Var[ X ]  E[X 2 ]   2
i.e., the variance is the mean of the squares
minus the square of the mean.
21 of 39
© Boardworks Ltd 2005
Variance and standard deviation
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.
We can write down the probability distribution of X:
x
P(X = x)
1
1
6
2
1
6
3
1
6
4
1
6
5
1
6
6
1
6
This probability distribution is symmetrical.
Therefore, E[X] = 3.5
22 of 39
© Boardworks Ltd 2005
Variance and standard deviation
To find the variance and standard deviation, we first need E[X2]:
x2
P(X = x)
So,
1
1
6
4
1
6
9
1
6
16
1
6
25
1
6
36
1
6
E[ X 2 ]  1 61    4  61    9  61   ...   36  61 
 15 61
Therefore, Var[ X ]  15   3
1
6

1 2
2
11
 2 12
So,   Var[ X ]  1.71 (to 3 s.f.)
23 of 39
© Boardworks Ltd 2005
Variance and standard deviation
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
P(X = x)
0
0.12
1
0.26
2
0.33
3
0.19
4
0.07
5
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.
24 of 39
© Boardworks Ltd 2005
Variance and standard deviation
x
P(X = x)
0
0.12
1
0.26
2
0.33
3
0.19
4
0.07
5
0.03
a) E[ X ]  (0  0.12)  (1 0.26)  ...  (5  0.03)  1.92
2
2
2
2
b) E[ X ]  (0  0.12)  (1  0.26)  ...  (5  0.03)  5.16
Therefore, Var[ X ]  5.16  1.922  1.4736
So,   1.4736  1.21 (3 s.f.)
25 of 39
© Boardworks Ltd 2005
Variance and standard deviation
c) Let Y be the random variable representing his weekly fine.
y
P(Y = y)
0
0.12
25
0.59
40
0.29
E[Y ]  (0  0.12)  (25  0.59)  (40  0.29)  26.35
So, his expected weekly fine is £26.35.
d) P(late exactly once in a 2 week period) =
(0.12  0.26)  (0.26  0.12)  0.0624
Not late
at all in
first week
26 of 39
Late once
in first
week
Late once
in second
week
Not late at
all in second
week
© Boardworks Ltd 2005
Variance and standard deviation
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.
27 of 39
© Boardworks Ltd 2005
Variance and standard deviation
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
b) The probability distribution of X is as follows:
0
x
P(X = x)
0.6
1
2
3
0.24 0.096 0.064
E[ X ]  (0  0.6)  (1 0.24)  (2  0.096)  (3  0.064)
 0.624
28 of 39
© Boardworks Ltd 2005
Variance and standard deviation
x
P(X = x)
0
0.6
1
0.24
2
0.096
3
0.064
E[ X 2 ]  (02  0.6)  (12  0.24)  (22  0.096)  (32  0.064)
 1.2
Using Var(X) = E[X2] – μ2, we get:
Var[ X ]  1.2  0.6242  0.810624
So, σ = 0.900 (to 3 s.f.)
29 of 39
© Boardworks Ltd 2005
Discrete uniform distribution
Contents
Introduction to discrete random variables
Cumulative distribution functions
Expectation
Variance and standard deviation for random
variables
Discrete uniform distribution
Expectation algebra – some key results
30 of 39
© Boardworks Ltd 2005
Discrete uniform distribution
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)
1
n
1
n
…
1
n
Examples of random variables that have
discrete uniform distributions are:
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
random digit generating program.
31 of 39
© Boardworks Ltd 2005
Discrete uniform distribution
An important special case is that of a discrete uniform
distribution taking the values 1, 2, …, n:
x
1
2
P(X = x)
1
n
1
n
…
…
n
1
n
This uniform distribution is sometimes denoted by U(1, n).
The mean and variance of this particular distribution are:
2
n
n

1
E[X] =
and
Var[X] =  1
12
2
32 of 39
© Boardworks Ltd 2005
Discrete uniform distribution
Example: A tetrahedral dice has its faces numbered 1, 2,
3 and 4. X is the score obtained when the dice is rolled.
X therefore has a uniform distribution, X ~ U(1, 4).
4 1
 2.5
So, E[X] =
2
42  1 5

and Var[X] =
12
4
33 of 39
© Boardworks Ltd 2005
Expectation algebra
Contents
Introduction to discrete random variables
Cumulative distribution functions
Expectation
Variance and standard deviation for random
variables
Discrete uniform distribution
Expectation algebra – some key results
34 of 39
© Boardworks Ltd 2005
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:
E[aX  b]   (axi  b) pi   (axi pi  bpi )
i
i
  axi pi   bpi  a  xi pi  b pi
i
i
i
i
 aE[ X ]  b
35 of 39
© Boardworks Ltd 2005
Expectation algebra
Result 2:
Var[aX + b] = a²Var[X]
Proof of result 2:
Var[aX  b]  E  (aX  b  E[aX  b]) 2 
 E  (aX  b  a   b)2 
Remember:
Var[ X ]  E[( X   )2 ]
where μ = E[X]
 E  (aX  a  )2   E  a 2 ( X   )2 
 a 2 Var[ X ]
36 of 39
© Boardworks Ltd 2005
Expectation algebra
Example: A random variable Y has mean 5 and variance 16.
Find the mean and the variance of:
a) 2Y
b) 3Y – 8
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
37 of 39
© Boardworks Ltd 2005
Expectation algebra
Examination-style question: A random variable X has
probability distribution given by:
c
P(X = x) =
for x = 1, 2, 3
x
a) Find c.
b) Find Var[X].
c) Find Var[6X + 2].
a) The probability distribution for X can be shown in a table:
x
P(X = x)
1
c
2
c
2
The sum of the probabilities
must be 1, so 1 56 c  1
38 of 39
3
c
3
6
i.e. c =
11
© Boardworks Ltd 2005
Expectation algebra
b) The probability distribution for X therefore, is:
1
6
11
x
P(X = x)
Using
2
3
11
3
2
11
6
E[ X ]  1 11
   2  113    3  112   1811
6
E[ X 2 ]  12  11
   22  113    32  112  
and
we get Var[ X ] 
36
11


18 2
11
36
11
 0.595 (to 3 s.f.)
c) Var[6X + 2] = 36 × 0.595 = 21.4 (to 3 s.f.)
39 of 39
© Boardworks Ltd 2005