Discrete Uniform Distribution
 Discrete
Uniform Distribution is
a probability distribution whereby a
finite number of equally spaced values
are equally likely to be observed;
every one of n values has equal
probability 1/n.
 A simple example of the discrete
uniform distribution is
fair die.
throwing a
Continuous Uniform Distribution
Continuous
Uniform
Distribution
or Rectangular Distribution is a family of
probability distributions such that for each
member of the family, all intervals of the same
length on the distribution's support are equally
probable.
Example -Uniform Distribution
The uniform distribution: all values are equally likely.
f(x)= 1, for 1 x 0
p(x)
1
1
x
We can see it’s a probability distribution because it
integrates to 1 (the area under the curve is 1):
1  x  1  0  1
1
0
1
0
Example: Uniform Distribution
What’s the probability that x is between 0 and ½?
p(x)
1
0
½
P(½ x 0)= ½
1
x
Expected Value and Variance
 All
probability distributions are
characterized by an expected value
(mean) and a variance (standard
deviation squared).
 Expected value is an extremely useful
concept for good decision-making!
Example: The Lottery
 A certain lottery works by picking 6
numbers from 1 to 49. It costs $1.00 to play
the lottery, and if you win, you win $2
million.
 If you play the lottery once, what are your
expected winnings or losses?
Lottery
Calculate the probability of winning in 1 try:
“49 choose 6”
1
 49 
 
 6

1
49 !

1
 7.2 x 10
-8
13 , 983 ,816
43 !6!
The probability function (note, sums to 1.0):
Out of 49 numbers,
this is the number of
distinct combinations
of 6.
x$
p(x)
-1
.999999928
+ 2 million
7.2 x 10--8
Expected Value
The probability function
x$
p(x)
-1
.999999928
+ 2 million
7.2 x 10--8
Expected Value
E(X) = P(win)*$2,000,000 + P(lose)*-$1.00
= 2.0 x 106 * 7.2 x 10-8+ .999999928 (-1) = .144 - .999999928
= -$.86
Negative expected value is never good!
You shouldn’t play if you expect to lose money!
Expected Value
If you play the lottery every week for 10 years,
what are your expected winnings or losses?
Expected Value
If you play the lottery every week for 10 years,
what are your expected winnings or losses?
520 x (-.86) = -$447.20
Expected Value of a Random
Variable
 Expected value is just the average or mean (µ)
of random variable x.
 It’s sometimes called a “weighted average”
because more frequent values of X are
weighted more highly in the average.
 It’s also how we expect X to behave on-average
over the long run (“frequentist” view again).
Expected Value
Discrete Case:
E(X ) 
x
i
p(x i )
all x
Continuous Case:
E(X ) 

x i p(x i )dx
all x
Symbol Interlude
 E(X) = µ
 These symbols are used interchangeably
Example: Expected Value
 Consider the following Probability Distribution:
x
10
11
12
13
14
P(x)
.4
.2
.2
.1
.1
5
x
i 1
i
p ( x )  10 (. 4 )  11 (. 2 )  12 (. 2 )  13 (. 1)  14 (. 1)  11 . 3
Example - Expectation
One thousand tickets are sold at $1
each for a colour television valued at
$500. What is the Expected Value of gain
if a person purchases one ticket ?
Example - Expectation
One thousand tickets are sold at $1
each for four prizes of $100, $50, $25 and
$10. What is the Expected Value of
winning a prize if a person purchases
two tickets ?
Variance/Standard Deviation
2 = Var(x) = E(x-)2
“The
expected (or average) squared distance
(or deviation) from the mean”
  Var ( x )  E [( x   ) ] 
2
2
 (x   )
i
all x
2
p(x i )
Variance
Discrete Case:
Var ( X ) 
 (x
  ) p(x i )
2
i
all x
Continuous Case:
Var ( X ) 

( x i   ) p(x i )dx
all x
2
Symbol Interlude
 Var(X)= 2
 SD(X) = 
 these symbols are used interchangeably
Practice Problem
It costs $1.00 to play the lottery.
The probability of winning the lottery is
18/38 and losing the lottery is 20/38. The
mean (already calculated)
What’s the variance of X ?
is -$.053.
Practice Problem

2

 (x
  ) p(x i )
2
i
all x
 (  1   . 053 ) (18 / 38 )  (  1   . 053 ) ( 20 / 38 )
2
2
 (1 . 053 ) (18 / 38 )  (  1  . 053 ) ( 20 / 38 )
2
2
 (1 . 053 ) (18 / 38 )  (  . 947 ) ( 20 / 38 )
2
2
 . 997
 
. 997  . 99
Standard deviation is $.99. Interpretation: On average, you’re
either 1 dollar above or 1 dollar below the mean, which is just under
zero. Makes sense!
Theorem
The variance of a random variable X is
б2 = E(X2) - µ2
OR
б2 = E(X2) – [E(X)]2
Example
Let the random variable X represent the number
of defective parts for a machine when 3 parts are
sampled from a production line and tested.
Calculate Variance of the following Probability
Distribution of X.
x
f(x)
0
0.51
1
0.38
2
0.10
3
0.01
Problem 1
Find the mean and the variance of
the random variable X with probability
function or density f(x):
f(x) = 2x, (0 ≤ x ≤ 1)
Problem 3
Find the mean and the variance of
the random variable X with probability
function f(x):
X = Number a fair die turns up
Problem 5
Find the mean and the variance of
the random variable X with probability
function or density f(x):
Uniform distribution on [ 0, 8 ]
Uniform Distribution
f(x) = 1/b-a
0
F(x) = x-a/b-a
0
a≤x≤b
Otherwise
a≤x≤b
Otherwise
Problem 7
What is the expected daily profit, if
a store sells X air conditioners per day
with probability f(10) = 0.1, f(11) = 0.3,
f(12) = 0.4, f(13) = 0.2, and profit per
air conditioner is $55 ?
Problem 9
If the mileage (in multiples of 1000
miles) after which a tyre must be replaced
is given by the random variable X with
density f(x) = θ e-θx (x > 0). What mileage
can you expect to get on one of these
tyres? Also find the probability that a tyre
will last at least 40,000 miles.
Problem 11
A small filling station is supplied with
gasoline
every
Saturday
afternoon.
Assume that its volume X of sales in ten
thousands of gallons has the probability
density f(x) = 6x ( 1 – x ), if 0 ≤ x ≤ 1 and 0
otherwise. Determine the mean, variance
and the standardized variable.
Standardized Random Variable
Z=X-µ
б
Std Rand Variable =
X - Mean
Std Deviation
Problem 13
Let X [cm] be the diameter of bolts in a
production. Assume that X has the density
f(x) = k ( x – 0.9 ) ( 1.1 – x ) if 0.9 < x < 1.1
and 0 otherwise. Determine k, sketch f(x),
and find µ and б2.
ASSIGNMENT NO 1
 Briefly describe following with the help of examples:
 Role of Probability
 Relation between Probability and Statistical
Inference
 Random Sampling
 Bayes’ Rule
 Joint Probability Distribution
 Hand Written assignments be submitted by 15 Nov 2012.
 Copying and Late Submissions will have appropriate
penalty.