Download Five Number Summaries

AS-Level Maths:
Statistics 1
for Edexcel
S1.5 Discrete random
variables
These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page.
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 = 1/9
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