Eighth lecture
Random
Variables
Consider the experiment of tossing a coin twice.
. If we are interested in the number of heads that
show on the top face, describe the sample space.
Solution:
S={ HH , HT , TH , TT }
2
1
1
0
w
S
X(w)
R
Definition (1):
A random variable is a function that associates a
real number with each element in the sample space.
Remark:
We shall use a capital letter, say X, to denote a
random variable and its corresponding small letter,
x in this case, for one of its values.
Example (1):
Two balls are drawn in succession without
replacement from an urn containing 4 red balls and 3
black balls. The possible outcomes and the values y of
the random variable Y, where Y is the number of red
balls, are
Sample Space
RR
RB
BR
BB
y
2
1
1
0
Example (2):
A stockroom clerk returns three safety helmets at random to
three steel mill employees who had previously checked them. If
Smith, Jones, and Brown, in that order, receive one of the three
hats, list the sample points for the possible orders of returning
the helmets, and find the value m of the random variable M
that represents the number of correct matches.
Solution:
If S, J and B stand for Smith’s,
Jones’s, and Brown’s helmets,
respectively, then the possible
arrangements in which the
helmets ma be returned and the
number of correct matches are
Sample Space
y
SJB
3
SBJ
1
BJS
1
JSB
1
JBS
0
BSJ
0
Example (3):
Interest centers around the proportion of people who
respond to a certain mail order solicitation. Let X be
that proportion. X is a random variable that takes on
all values x for which 0 ≤ x ≤ 1.
Example (4):
Let X be the random variable defined by the waiting
time, in hours, between successive speeders spotted by
a rader unit. The random variable X takes on all
values x for which x≥ 0.
Definition (2):
If a sample space contains a finite number of
possibilities or an unending sequence with as many
elements as there are whole numbers , it is called a
discrete sample space.
Definition (3):
If a sample space contains an infinite number of
possibilities equal to the number of points on a line
segment, it is called a continuous sample space.
Types of random variables:
1. Discrete random variable.
A random variable is called a discrete random variable if its set
of possible outcomes is countable.
2. Continuous random variable.
A random variable is called a continuous random variable when
can take on values on a continuous scale .
Example (5):
Classify the following random variables as discrete or continuous:
X: the number of automobile accidents per year in Virginia.
Y: the length of time to play 18 holes of golf.
M: the amount of milk produced yearly by a particular cow.
N: the number of eggs laid each month by a hen.
P: the number of building permits issued each month in a certain
city.
Q: the weight of grain produced per acre.
Definition (4):
The set of ordered pairs (x, f(x)) is a probability function,
probability mass function, or probability distribution of the
discrete random variable X if, for each possible outcome x,
1  f ( x )  0,
2
f
( x )  1,
x
3  P ( X  x )  f ( x ).
Example(6):
Determine the value c so that each of the following function can
serve as a probability distribution of the discrete random
variable X:
f(x)=c(x2+4), for x= 0,1,2,3.
Example(7):
Let W be a random variable giving the number of heads minus
the number of tails in three tosses of a coin. List the elements of
the sample space S for the three tosses of the coin and to each
sample points assign a value w of W.
Example(8):
Find a formula for the probability distribution of the random
variable X representing the outcome when a single die is rolled
once.
Definition (5):
The cumulative distribution function F(x) of a discrete random
variable X with probability distribution f(x) is
F (x )  P (X  x ) 
f
(t ), for    x  
t x
Example(9):
If
4 
 
x 
, x  0,1, 2, 3, 4 :
f (x ) 
16
Prove that f(x) is a probability mass function?
Find the cumulative distribution function of the random variable
X.
The probability that X assumes a
value between a and b is equal to the
shaded area under the density
function between the ordinates at x=a
and x=b and from integral calculus is
given
P (a  X  b ) 

f(x)
b
a
f ( x )d x
a
b
x
Definition (6):
The function f(x) is a probability density function for the
continuous random variable X, defined over the set of real
number R, if
1  f ( x )  0, for all x  R
2



f ( x )d x  1
3  P (a  X  b ) 

b
f ( x )d x .
a
Example(10):
Suppose that the error in the reaction temperature, in oC, for
controlled laboratory experiment is a continuous random
variable X having the probability density function
x 2
(a) Verify condition (2) of Definition(6).
, 1  x  2

f (x )   3
(b) Find P(0<X≤1)
 0, elsew h er e

Definition (7):
The cumulative distribution function F(x) of a continuous
random variable X with density function f(x) is
F (x )  P (X  x ) 

x

f (t )d t , fo r    x  
As an immediate consequence of Definition (7) one can
write the two results,
P ( a  X  b )  F (b )  F ( a ), an d f ( x ) 
d F (x )
dx
If the derivative exists.
Example(11):
For the density function of Example (10) find F(x) and use it to
evaluate P(0<X≤1)
x 2
, 1  x  2

f (x )   3
 0, elsew h er e

Example(12):
The waiting time, in hours, between successive speeders
spotted by a radar unit is a continuous random variable
with cumulative distribution function
 0, x  0
F (x )  
8 x
1

e
,x  0

Find the probability of waiting less than 12 minutes between
successive speeders
(a)Using the cumulative distribution function of X;
(b) Using the probability density function of X.
Definition(8):
Let X a random variable with probability distribution f(x).
The mean or expected value of X is
  E (x ) 
x
f (x )
x
If X is discrete, and
  E (x ) 



x f ( x )d x
If X is continuous.
Example (13):
A lot containing 7 components is sampled by a quality
inspector; the lot contains 4 good components and 3 defective
components. A sample of 3 is taken by the inspector. Find the
expected value of the number of good components in this
sample.
Example(14):
Let X be the random variable that denotes the life in hours
of a certain electronic device. The probability density
function is
 20, 000
, x  100

3
f (x )   x
 0, elsew h er e

Find the expected life of this type of device.
Definition(9):
Let X be a random variable with probability distribution
f(x). The expected value of the random variable g(x) is
 g (X )  E  g (X ) 
 g (x ) f
(x )
x
If X is discrete, and
 g (X )  E  g (X ) 
If X is continuous.



g ( x ) f ( x )d x
Example (15):
Suppose that the number of cars X that pass through a car
wash between 4 p.m. and 5 p.m. on any sunny Friday has the
following probability distribution:
x
P(X=x)
4
5
1/12 1/12
6
1/4
7
1/4
8
1/6
9
1/6
Let g(X)=2X-1 represent the amount of money in dollars, paid
to the attendant by the manager. Find the attendant’s expected
earning for this particular time period.
Definition(8):
Let X a random variable with probability distribution f(x)
and mean . The variance of X is
2
2
2
s  E (X   )  
( x   ) f ( x ) if X is discrete, and

s
2

2
 E  ( X   )  

x



(x   ) f (x )
2
if X is continuous.
The positive square root of the variance, s, is called
standard deviation of X.
Example(16):
Suppose that the number of cars X that pass through a car
wash between 4 p.m. and 5 p.m. on any sunny Friday has the
following probability distribution:
x
P(X=x)
4
5
1/12 1/12
Find the variance of X.
6
1/4
7
1/4
8
1/6
9
1/6
Theorem(1):
The variance of a random variable X is
s
2
 E (X )  
2
2
Example(16):
Use Theorem (1) to find the variance in Example (15).
Example(17):
The weekly demand for Pepsi, in thousand of liters, from a
local chain of efficiency stores, is a continuous random
variable X having the probability density
 2( x  1),1  x  2
f (x )  
 0, elesw h er e
Find the mean and the variance of X.
Theorem(2):
If a and b are conestants, then
E(aX+b)=aE(X)+b
Corollary(1):
Setting a=0, we see that E(b)=b
Corollary(2):
Setting b=0, we see that E(aX)=aE(X)
Theorem(3):
If a and b are conestants, then
2
2 s2
s
=a
aX+b
X
Corollary(1):
Setting a=1, we see that s2X+b=s2X
Corollary(2):
Setting b=0, we see that s2aX=a2s2X
Example(18):
Use the result above to Find the mean of 2X-1 in Example(18).
x
P(X=x)
4
5
1/12 1/12
6
1/4
7
1/4
8
1/6
9
1/6