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Variance of Probability Distribution
Spread
Standard Deviation
Unbiased Estimate
Sample Variance and Standard Deviation
Alternative Definitions
Chebychev's Inequality
1


Let X be a random variable with values x1, x2,
…, xN and respective probabilities p1, p2,…, pN.
The variance of the probability distribution is
 variance   x1   
2
p1   x2    p2 
2
  xN    p N .
2
2

Roughly speaking, the variance measures the
dispersal or spread of a distribution about its
mean. The probability distribution whose
histogram is drawn on the left has a smaller
variance than that on the right.
3

The standard deviation of probability distribution is
  2 
 variance, where
   x1    p1   x2    p2 
2
or 
2
2
x1   


2
2
f1   x2    f 2 
2
N
  xr    pr ,
2
  xr    f r
2
.
4

Compute the
variance and the
standard deviation
for the population of
scores on a fivequestion quiz in the
table.

0  4   1 9   2  6   3 14   4 18   5  9 
60
3
5
132
Variance   
 2.2,   2.2  1.48
60
2
6

If the average of a statistic, if that statistic were
computed for each sample, equals the
associated parameter for the population, then
that statistic is said to be unbiased.
7

s2

The unbiased variance for a sample is
x  x


1
2

f1  x2  x

2
f2 
n 1

 xr  x

2
fr
.
The unbiased standard deviation for a sample
is
s s .
2
8

Compute the
sample variance
and standard
deviation for the
weekly sales of
car dealership A.
9
x
5  2   6  2   7 13  8  20   9 10   10  4   111
52
 7.96
s 2  511 [ 5  7.96   2   6  7.96   2   7  7.96  13
2
2
2
  8  7.96   20   9  7.96  10  10  7.96   4
2
2
2
 11  7.96  1]  1.45
2
s  1.45  1.20
10

Two alternative definitions for variance are
 2  E  X 2   2

and, for a binomial random variable with
parameters n, p, and q,
 2  npq.
11

Find the variance when a fair coin is tossed 5
times and X is the number of heads.
 1  1  5
  5    
 2  2  4
2
12

Chebychev's Inequality Suppose that a
probability distribution with numerical
outcomes has expected value  and standard
deviation  . Then the probability that a
randomly chosen outcome lies between  - c
and  + c is at least
 c .
1 
2
13

A drug company sells bottles containing 100
capsules of penicillin. Due to bottling
procedure, not every bottle contains exactly 100
capsules. Assume that the average number of
capsules in a bottle is 100 and the standard
deviation is 2. If the company ships 5000
bottles, estimate the number of bottles having
between 95 and 105 capsules, inclusive.
14

The number of bottles containing between 100 - 5
and
100 + 5 capsules can be estimated using
Chebychev's Inequality.
  100
 2
c5
 c
1 
2
 5
 1 2
2
 .84
On average, at least 84% will contain between 95 and
105 capsules. 4200 bottles.
15


The variance of a random variable is the sum
of the products of the square of each
outcome's distance from the expected value
and the outcome's probability. The variance of
the random variable X can also be computed
as
E(X 2) - [E(X)] 2.
A binomial random variable with parameters
n and p has expected value np and variance
np(1 - p).
16


The square root of the variance is called the
standard deviation.
Chebychev's Inequality states that the
probability that an outcome of an experiment is
within c units of the mean is at least
2 ,

1

where is the standard deviation.
c

 
17