Transcript S1.5 Discrete random variables

AS-Level Maths:
Statistics 1
for Edexcel
S1.5 Discrete random
variables
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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
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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.
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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.
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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.
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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 ,
So,
i
Therefore
c + 2c + 3c + 2c + c = 1
9c = 1
c = 91
2 1 1
P(Y > 2) = P(Y = 3 or 4) = 2c + c =  
9 9 3
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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.
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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)
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1
5
7
2
5
21
3
1
21
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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
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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)
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5
10 15 20 25
0.1 0.3 0.5 0.8 0.9
30
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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
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2
4k
3
4
9k 16k
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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
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Gambling games
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Gambling games
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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
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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.
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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).
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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.
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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
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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
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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.
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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
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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.)
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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.
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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.)
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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
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Late once
in first
week
Late once
in second
week
Not late at
all in second
week
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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.
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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
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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.)
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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
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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.
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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
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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
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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
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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
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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 ]
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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
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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
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3
c
3
6
i.e. c =
11
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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.)
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