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```Probability Distributions – Finite RV’s
• Random variables first introduced in Expected
Value
• def. A finite random variable is a random variable
that can assume only a finite number of distinct
values
• Example: Experiment-Toss a fair coin twice
X( random variable)- number of
X can assume only 0, 1, 2
Probability mass function
def. The probability mass function (p.m.f.)
of a finite random variable, X, is given by
f X ( x)  P( X  x) .
 0  f X ( x )  1.
  f X ( x)  1.
all x
Probability mass function(p.m.f)Small f
Example 1: Box contains four \$1 chips,
three \$5 chips, two \$25 chips, and one \$100
chip. Let X be the denomination of a chip
selected at random. The p.m.f. of X is
displayed below.
x
f X (x )
\$1
0.4
\$5
0.3
\$25
0.2
\$100
0.1
fX (x )
Probability Mass Function
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
\$1
\$5
\$25
x
\$100
Cumulative distribution
function(c.d.f)
def. The cumulative distribution function
(c.d.f.) of any random variable, X, is given
by FX ( x )  P ( X  x ) .
Domain consists of all real numbers
 FX ( x )  0 as x  
 FX ( x )  1 as x  
Cumulative distribution
function(c.d.f)- Big F
Example 1 (continued): The c.d.f. of X is
displayed below.
0
0.4

FX ( x)  0.7
0.9

1.0
if x  1
if 1  x  5
if 5  x  25
if 25  x  100
if x  100
Cumulative Distribution Function
1.2
1.0
FX (x )
0.8
0.6
0.4
0.2
0.0
-20
0
20
40
60
x
80
100
120
Calculating Probabilities-Using
p.m.f & c.d.f
P( X  \$25)  f X (\$25)  0.2
P ( X  \$25)  P ( X  \$1)  P ( X  \$5)  P ( X  \$25)
 f X (\$1)  f X (\$5)  f X (\$25)
 0.4  0.3  0.2
 0.9
P( X  \$25)  FX (\$25)  0.9
P ( X  \$5)  f X (\$5)  f X (\$25)  f X (\$100)
 0. 3  0 . 2  0. 1
 0. 6
P( X  \$5)  1  P( X  \$5)
 1  P( X  \$1)
 1  FX (\$1)
 1  0.4
 0.6
Example 2: The c.d.f. of a random variable
X is given below.
0
0.1

0.3
FX ( x)  
0.7
0.9

1.0
Find the p.m.f. of X.
if x  0
if 0  x  1
if 1  x  4
if 4  x  9
if 9  x  16
if x  16
p.m.f
x
f X (x )
0
0.1
1
0.2
4
0.4
9
0.2
16
0.1
Expected value –Finite R.V
The expected value or mean of a finite
random variable is given by
E( X )  μX 
 x  f X ( x) .
all x
Example 1 (continued):
x
\$ 1
\$ 5
\$ 25
\$ 100
Sum
f X (x)
x  f X (x)
0.4
0.40
0.3
1.50
0.2
5.00
0.1
10.00
1.0  X  \$16.90
E( X )   X
 \$1 0.4  \$5  0.3  \$25  0.2  \$100  0.1
 \$16.90
Example 3: A recent Gallup Poll showed
that 55% of Americans own a cell phone.
Let X be the number of Americans in a
sample of size three who own a cell phone.
Let C be the event than an individual owns a
cell phone and let D be the event that an
individual does not own a cell phone.
CCC CCD CDC CDD 
S 

DCC
DCD
DDC
DDD


P(C )  0.55 and P( D)  1  0.55  0.45
P (CCD )  P (C  C  D)
 P (C )  P (C )  P ( D )
 (0.55) 2  (0.45)
 0.136125
P(CCD )  P(CDC )  P( DCC )  0.136125
P( X  2 )  f X ( 2 )
 P( CCD  CDC  DCC )
 P( CCD )  P( CDC )  P( DCC )
 0.136125  0.136125  0.136125
 0.408375
The p.m.f. of X is given by
0
1
2
3
x
f X (x ) 0.091125 0.334125 0.408375 0.166375
The c.d.f. is given by
0
0.091125

FX ( x)  0.425250
0.833625

1
if
if
if
if
if
x0
0  x 1
1 x  2
2 x3
x3
The random variable in this example is a
special kind of finite random variable.
A Bernoulli trial is an experiment that has
exactly two outcomes, “success” and
“failure”.
A binomial random variable gives the
number of “successes” in n independent
Bernoulli trials, where the probability of
“success” on each trial is equal to p.
BINOMDIST-show excel
The expected value of a binomial random
variable, X, with parameters n and p is given
by E ( X )   X  n  p .
Example 3 (continued-call problem):
E ( X )   X  3  0.55  1.65 people
Probability Distributions –
Continuous RV’s
def. A continuous random variable is a random
variable than can assume any value in some interval
of numbers.
Examples: T-the length of time, measured in minutes,
and parts of minutes,between arrivals of phone calls
at a company switch board
In theory-T can assume any positive real number as a
value
Here the interval-
[ 0, )
Probability density function-p.d.f
def. The probability density function (p.d.f.)
of a continuous random variable, X, is given
by f X (x ) .
 f X ( x )  0 .
The total area between the graph of
f X (x ) and the horizontal axis must be
equal to 1.
p.d.f
Example 4: The p.d.f. of T, the weekly CPU
time (in hours) used by an accounting firm,
is given below.
if t  0
0
3 2
fT (t )   t (4  t ) if 0  t  4
64
0
if t  4

0.5
0.4
0.3
f T (t )
0.2
0.1
0
-4
-2
-0.1
0
2
t
4
6
Relationship between Probability
& Area of p.d.f - for Continuous
R.V
P(a  X  b) is equal to the area between
the graph of f X (x ) and the horizontal axis
over the interval [a, b] .
Example 4 (continued): P(1  T  2) is equal
to the area between the graph of fT (t ) and
the t-axis over the interval [1,2] .
0.5
0.4
0.3
f T (t )
0.2
0.1
0
-4
-2
-0.1
0
2
t
4
6
Important
For any continuous random X,
P( X  x)  0 .
Example 4 (continued): The c.d.f. of T is
given below.
if t  0
0
 1 3
FT (t )   t (16  3t ) if 0  t  4
256
1
if t  4

c.d.f
1.2
1.0
0.8
0.6
F T (t )
0.4
0.2
0.0
-4
-2
-0.2 0
2
t
4
6
Use the c.d.f. to find P(1  T  2) .
P( 1  T  2 )  P( T  2 )  P( T  1 )
 P( T  2 )  P( T  1 )
 FT ( 2 )  FT ( 1 )
1
1 3
3

( 2 )( 16  3  2 ) 
( 1 )( 16  3 1 )
256
256
 0.2617
Additional tools are needed to compute the
expected value of a continuous random
variable.
Uniform random variable
def. A continuous uniform random variable
is a random variable defined on an interval
[a, b] such that every subinterval of [ a, b]
having the same length has the same
probability.
p.d.f for uniform random variable
If X is a continuous uniform random
variable on the interval [ 0,u ] , then
 0 if x  0
1
fX ( x )  
if 0  x  u
u
 0 if x  u .
c.d.f for uniform random variable
 0 if x  0
x
FX ( x)  
if 0  x  u
u

1
if
x

u
.

Expected value for uniform
random variable
u
E( X )   X 
2
Example for Uniform random
variable
Example 5: A bus arrives at a bus stop every
10 minutes. Let W be the waiting time (in
minutes) until the next bus. The p.d.f. and
c.d.f. of W are given below.
0 if w  0
1
fW ( w)   if 0  w  10
10
0 if w  10
Graph of p.d.f for uniform
0.120
0.100
0.080
f W (w )
0.060
0.040
0.020
0.000
-5
-0.020 0
5
w
10
15
Graph of c.d.f for uniform
0 if w  0
w
FW ( w)  
if 0  w  10
10
if w  10
1
1.2
1.0
0.8
F W (w )
0.6
0.4
0.2
0.0
-5
-0.2 0
5
w
10
15
Find P(4  W  6) .
0.120
0.100
0.080
f W (w )
0.060
0.040
0.020
0.000
-5
-0.020 0
5
w
10
15
P(4  W  6)  (6  4)  0.10  0.2
P (4  W  6)  P (W  6)  P (W  4)
 P (W  6)  P (W  4)
 FW (6)  FW (4)
6 4
 
10 10
 0.2
Example 5 (continued): The expected value
10
E (W )  W 
5
of W is given by
.
2
Exponential random variable
def. An exponential random variable may be
used to model the length of time between
consecutive occurrences of some event in a
fixed unit of space or time.
p.d.f/c.d.f/ expected value –
Exponential random variable
If X is an exponential random variable with
parameter  , then
0
f X ( x)   1  x / 
 e
0
FX ( x )  
x /
1  e
E( X )   X   .
if x  0
if x  0.
if x  0
if x  0.
Example 6: On average, three customers per
hour use the ATM in a local grocery store.
Let T be the time (in minutes) between
consecutive customers. The p.d.f. and c.d.f.
of T are given below.
if t  0
0
fT (t )   1 t / 20
if t  0
 20 e
0.06
0.05
0.04
f T (t )
0.03
0.02
0.01
0.00
-20 -0.01 0
20
40
60
t
80
100
120
if t  0
0
FT (t )  
t / 20
1

e
if t  0

1.2
1.0
0.8
F T (t )
0.6
0.4
0.2
0.0
-20 -0.2 0
20
40
60
t
80
100
120
Find P(T  15) .
0.06
0.05
0.04
f T (t )
0.03
0.02
0.01
0.00
-20 -0.01 0
20
40
60
t
P (T  15)  1  P(T  15)
 1  P(T  15)
 1  (1  e 15 / 20 )
 0.4724
80
100
120
Example 6 (continued): The expected value
of T is given by E (T )  T    20 .
```
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