Download Chapter 2: Random variables

ROHANA
BINTI
ABDUL
HAMID
MADAM
ROHANA
BINTI
ABDUL HAMID
INSTITUT E FOR ENGINEERING MATHEMATICS (IMK)
INSTITUT EMALAYSIA
FOR ENGINEERING
UNIVERSITI
PERLIS MATHEMATICS (IMK)
UNIVERSITI MALAYSIA PERLIS
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2.1Introduction
2.2Discrete probability
distribution
2.3Continuous probability
distribution
2.4Cumulative distribution
function
2.5Expected value, variance and
standard deviation

In an experiment of chance, outcomes occur
randomly. We often summarize the outcome
from a random experiment by a simple number.
Definition 2.1
 A variable is a symbol such as X, Y, Z, x or H,
that assumes values for different elements. If
the variable can assume only one value, it is
called a constant.

A random variable is a variable whose value is
determined by the outcome of a random
experiment.
Example 2.1
A balanced coin is tossed two times. List the elements of
the sample space, the corresponding probabilities and
the corresponding values X, where X is the number of
getting head.
Solution
Elements of
sample space
HH
HT
TH
TT
Probability
X
¼
¼
¼
¼
2
1
1
0
TWO TYPES OF RANDOM
VARIABLES
Discrete
Random
Variables
A random
variable is
discrete if its
set of possible
values consist
of discrete
points on the
number line.
Continuous
Random
Variables
A random
variable is
continuous if
its set of
possible values
consist of an
entire interval
on the number
line.
2.2 DISCRETE PROBABILITY
DISTRIBUTIONS
Definition 2.3:
 If X is a discrete random variable, the function
given by f(x)=P(X=x) for each x within the range
of X is called the probability distribution of X.

Requirements for a discrete probability
distribution:
1) The probability of each value of the discrete random variable
is between 0 and 1, inclusive. That is, 0  P( X  x)  1
2) The sum of all the probabilities is 1. That is,
 P( X  x)  1
xS
Example 2.2
 Check whether the distribution is a probability
distribution.
x
0
1
2
3
4
P(X=x)
0.125
0.375
0.025
0.375
0.125
Solution
All
the probabilities between 0 and 1.
4
2. 
. P( X  x)  P( X  0)  P( X  1)  P( X  2)  P( X  3)  P( X  4)
1.
0
= 0.125+0.375+0.025+0.375+0.125
= 1.025
1
 so the distribution is not a probability distribution.
Example 2.3
 Check whether the function given by
x2
f ( x) 
for x =1,2,3,4,5
25
can serve as the probability distribution of a discrete
random variable.
Solution
1) The probabilities are between 0 and 1
f(1) = 3/25 , f(2) = 4/25, f(3) =5/25, f(4) = 6/25, f(5)=7/25
Solution
5

1
x2
f ( x)  
25
1
= f (1)  f (2)  f (3)  f (4)  f (5)
1 2 2  2 3  2 4  2 5  2
=




25
25
25
25
25
3
4
5
6
7
=
   
25 25 25 25 25
25
=
25
1
5
# so the given function is a probability distribution of a
discrete random variable.
2.3 CONTINUOUS PROBABILITY
DISTRIBUTIONS
Definition 2.4:
 A function with values f(x), defined over the set
of all numbers, is called a probability density
function of the continuous random variable X if
and only if
b
P(a  X  b)   f ( x) dx
a
for any real constant a and b with a  b
Requirements for a probability density function of
a continuous random variable X:
1) f ( x)  0 for -  x  
2)



f ( x) dx  1. That is the total area under graph is 1.
Example 2.4
Let X be a continuous random variable with the
following
3 2
 ( x  1), 0  x  1
f ( x)   4
0,
otherwise
a) Verify whether this distribution is a probability
density function
b) Find P (0  X  0.5)
c)
Find P(0.5  X  2)
Answer;
The distribution is probability density function if it fulfill the
following requirements,
1) All f(x)≥0
2) If 
a)
 f ( x)dx  1

In this problem,
1) First requirement
f(0)=3/4≥0, f(1)=3/2≥0, f(x)=0, otherwise
- All f(x)≥0
Must write the conclusion
so that we know the first
requirement is fulfill!
2) Second requirement
0
1

3 2
0dx  0 4 ( x  1)dx  1 0dx
3 x3
 [  x]10
4 3
3 4
 [ ]
4 3
1
-
Must write the conclusion
so that we know the
second requirement is
fulfill!

 f ( x)dx  1

Write last
conclusion to answer
the question!
Since all requirements all fulfill, the distribution is probability density
function.
b) P(0  X  0.5)
0.5
3 2
  ( x  1)dx
4
0
0.5
3
  ( x 2  1)dx
40
0 .5
3

3 x
   x
4 3
0
3


3 0.5
 (
 0.5)  0
4 3

 0.40625
c) P(0.5  X  2)
1
2
3 2
  ( x  1)dx   0dx
4
0.5
1
1
3
2
  ( x  1)dx  0
4 0.5
1
3

3 x
   x
4 3
 0 .5

3 1
0.5
 (  1)  (
 0.5)
4 3
3

3
3
 0.59375
2.4 CUMULATIVE DISTRIBUTION
FUNCTION

The cumulative distribution function of a discrete
random variable X , denoted as F(X), is
F ( x)  P( X  x)   f (t ) for    x  
tx

For a discrete random variable X, F(x) satisfies the
following properties:
1) 0  F ( x)  1
2) If x  y, then F ( x)  F ( y )

If the range of a random variable X consists of the
values x1  x2  x3  ...  xn , then f ( x1 )  F ( x1 ) and
f ( xi )  F ( xi )  F ( xi 1 ) for i  2, 3,..., n

The cumulative distribution function of a
continuous random variable X is
x
F ( x)  P( X  x) 

f (t ) dt for    x  

Let F  x  be the distribution function for a continuous random
dF ( x)
variable X . Then f ( x) 
 F ( x)
dx
wherever the derivative exists.
Example 2.5 (Discrete random variable)
5 x
Given the probability function f ( x) 
for x  1, 2,3, 4,
10
find F ( x)
Solution
x
1
2
3
4
f(x)
4/10
3/10
2/10
1/10
F(x)
4/10
7/10
9/10
1
Example 2.6 a.(Continuous random variable)
Given the probability density function of a random
variable X as follows;
3 2
 x ,
f ( x)   8
0,
0 x2
otherwise
Find the cumulative distribution function, F(X)
ii) Find P(1  X  2)
i)
Solution
i) Cummulative distribution function, F(X)
i) For x  0 ,
x
F ( X )   0dt  0

ii)
For 0  x  2
0
x
3 2
F ( X )   0dt   t dt
8

0
3
t

8
3
x
0
x

8
For x  2 ,
0
2
x
3 2
F ( X )   0dt   t dt   0dt
8

0
2
t3

8
2
0
0,
 3
x
F(X )   ,
8
1,
23

1
8
  x  0
0 x2
x2
Summary is
important!!!
Two methods to answer question 3
iii ) P (1  X  2)
 F (2)  F (1)
23 13
 
8 8
7

8
iii) P (1  X  2)
2

3 2

x dx
8
1
3
3 x 2
 [ ]1
8 3
3
3
3 2 1
 [  ]
8 3 3
7

8
TRY!!!
 The lifetime of a computer, in x years, is given by
a random variable with the following function.
x
4 ,

4  x
f ( x)  
,
 4
0


0 x2
2 x4
elsewhere
a) Verify that the function is probability density
function.
b) Find the cumulative distribution function of the
random variable X.
c) Find P(1≤X≤3)
2.5 EXPECTED VALUE, VARIANCE AND
STANDARD DEVIATION
2.5.1 Expected Value


The mean of a random variable X is also known
as the expected value of X as    X  E ( X )
If X is a discrete
random variable,
If X is a continuous
random variable,
2.5.2 Variance
Var ( X )   2   X2  E (( X   ) 2 ), where
Var ( X )   ( X   ) 2 P ( X  x), in the discrete case,
xS

Var ( X ) 
2
(
X


)
f ( x)dx , in the continuous case when it exists.


Var ( X ) exists if and only if   E ( X ) and E ( X 2 ) both exist, and
then Var ( X )  E ( X 2 )  ( E ( X )) 2
2.5.3 Standard Deviation
The standard deviation is    X  Var ( X )   X2
2.5.4 Properties of Expected Values
• For any constant a and b,
i) E (a)  a
ii) E (bX )  bE ( X )
iii) E (aX  b)  aE ( X )  b
2.5.5 Properties of Variances
For any constant a and b,
i) Var (a )  0
ii) Var (bX )  b Var ( X )
2
iii) Var (aX  b)  a 2Var ( X )
Example 2.7
Find the mean, variance and standard deviation
of the probability function
x
f ( x) 
for x  1, 2,3, 4
10
Solution
Mean:
n
E  X    x  f  x
i 1
4
x
 x
10
i 1
1
2
3
4
 1  2   3   4 
10
10
10
10
30

3
10
Varians:
E  X 2    x2  f  x 
n
i 1
4
x
x 
10
i 1
2
3
2 1
2
2
2 4
1   2  3   4 
10
10
10
10
 10
2
Var ( X )  E ( X 2 )  ( E ( X )) 2
 10  32  1
 X  Var ( X )  1
Example 2.8
Let X be a continuous random variable with the
following probability density function
3
 x(2  x), 0  x  2
f ( x)   4
0
, otherwise
Find
a) E ( X )
b) Var ( X )
Answer;

a) E (X) =  x. f ( x)dx


3

E ( X )   x.0dx   x. x(2  x)dx   x.0dx
4


0
2
0
2
2
3
  x 2 .(2  x)dx
40
3 2 x3 x 4 2
 [
 ]0
4 3
4
3 4
 . 1
4 3
b) Var (X) = E (X2) – (E(X))2
E (X) = 1,


2
2
2 3
E ( X )   x .0dx   x . x(2  x)dx   x 2 .0dx
4


0
2
0
2
2
3
  x 3 .(2  x)dx
40
3 2 x 4 x5 2
 [
 ]0
4 4
5
3 8 6
 . 
4 5 5
6
1
2
Var ( X )   (1) 
5
5
Exercise 1
A survey asked a sample of people living in Kangar,
how many times they shop at The Store supermarket
each week and the following distribution is obtained;
X
0
1
2
3
4
P(X=x)
0.2
0.35
0.25
0.11
0.09
Does the distribution determine the probability distribution
for discrete random variable?
b) Find the probability of people shop at The Store more than
2 times.
(answer: 0.2)
c) Find the mean and variance of number of times people
living in Kangar shop at The store supermarket.
a)
(mean=1.54, variance=1.4084)
Exercise 2
A restaurant manager is considering a new location for
her restaurant. The projected annual cash flow for the
new location is;
Annual
cash flow
Probability
$10,000
$30,000
$70,000
$90,000
$100,000
0.10
0.15
0.50
0.15
?????
Given that the distribution is probability distribution,
find the expected cash flow for the new location.
(E(X)=64 000)
Exercise 3
X is a random variable which represents the delay time,
in minutes, for a Star Cruise to depart from Pulau
Langkawi jetty, with function
0.1  0.005 x,
f ( x)  
0,
0  x  20
otherwise
a) Verify that the function is probability density
function.
b) Calculate F (X).
c) Find E(X) and Var (X).
d) Find P(10≤X≤30)