
Consider a university having 15,000 students
and let X=number of courses for which a
randomly selected student is registered. The
pmf of X is below. Since p(1)=.01, we know
that (.01)(15,000)=150 students are registered
for one course, and similarly for other numbers
of courses.
x
1
2
3
4
5
6
7
p(x)
.01
.03
.13
.25
.39
.17
.02
#
150
450
1950
3750
5850
2550
300
2

To obtain the average number of courses per
student, we compute the total number of
courses taken and divide the total number of
students.
1(150)  2(450)  3(1950)   7(300)
 4.57
15,000

Since 150/15,000=.01=p(1), 450/15,000 =.03=p(2)
etc., the formula for the average value could also
be written as
1 p(1)  2  p(2)   7  p(7)  4.57
3

Thus, to compute the population average
value, we only need the possible values for X
and the probabilities of X taking on those
values.
4

Let X be a discrete rv with the set of
possible values D and pmf p(x). The
expected value (or mean) of X, denoted by
E(X) or  , is
E ( X )     xD x  p( x)
5
x
0
1
2
3
4
5
6
p(x)
.05
.10
.15
.25
.20
.15
.10
E ( X )  0(.05)  1(.10)  2(.15)  3(.25)  4(.20)  5(.15)  6(.10)  3.3
6
E ( X )  p 1  (1  p)  0  p
7

If the rv X has a set of possible values D and
pmf p(x), then the expected value of any
function h(X) is computed by
E  h  X     xD h  x   p  x 
8

A computer store purchases three computers
for $500 apiece. The manufacturer agrees to
repurchase any unsold computers after a
specified period at $200 each. Let X denote
the number of computers sold, and supposed
that p(0)=.1, p(1)=.2, p(2)=.3, p(3)=.4. Let h(X)
denote the profit associated with selling X
units. If the computers are sold for $1000,
what is the expected profit?
9
h( X )  1000 X  200(3  X )  1500  800 X  900
E  h( X )  E 800 X  900   900 .1   100 .2 
  700 .3  1500 .4   $700
10
E  aX  b   aE  X   b

Proof:
E  aX  b    xD  ax  b  p  x 
 a  xD x  p  x   b xD p  x 
 aE  X   b

Two special cases are E  aX   aE  X  and
E  X  b  E  X   b
11

p(0)=.1, p(1)=.2, p(2)=.3, p(3)=.4
E  X   1.2   2 .3   3.4   2
E 800 X  900   800 E  X   900  $700
12

The expected value of X is a measure of
where the probability mass is centered. Two
distributions can have the same expected
value, but the probability mass can be spread
differently.

Variance gives a measure of spread about the
mean.
13

Let X have pmf p(x) and expected value  .
The variance of X,  2 , is
V ( X )    E  X      xD  x     p  x 
2

2
2
The standard deviation of X is
 
2
14

It is easier to use the following formula to
compute the variance


2
2
  E  X    E  X      x  p  x    
 xD

2

2
2
Proof:    xD  x     p  x 
2
2
  xD x 2  p  x   2  xD x  p  x    2  xD p  x 
 E X 2   2
15
V  aX  b   a 2V  x 

Proof: V  aX  b   E  aX  b     E  aX  b  
2
2
 E  a 2 X 2  2abX  b 2    a 2  2  2ab  b 2 
2
2
2
2 2
2



  a E  X   2ab  b    a   2ab  b 
 a V  x
2
16

p(0)=.1, p(1)=.2, p(2)=.3, p(3)=.4
V  X   1
2
 .2   2 .3   3  .4   2
2
2
2
1
V  800 X  900   8002V  X   $640,000
17