Download MTH 245: Mathematics for Management, Life, and Social Sciences

Section 7.5: The Variance and Standard Deviation.
MTH 245: Mathematics for Management, Life,
and Social Sciences
F. Patricia Medina
Department of Mathematics. Oregon State University
November 9, 2014
Section 7.5
Section 7.5: The Variance and Standard Deviation.
The Variance and Standard Deviation.
We will now define the variance of our sample/population/random
variable. This is meant to tell us how spread-out our data or
distributions are.
Definition 1
The variance of a random variable X is the expected value of its
squared deviation from µ = E(X). If X takes values x1 , x2 , . . . , xn and
probabilities p1 , p2 , . . . , pn (probability distribution),
Var(X) = (x1 − µ)2 · p1 + (x2 − µ)2 · p2 + · · · + (xN − µ)2 · pN .
Section 7.5: The Variance and Standard Deviation.
Example 2
Compute the expected value and variance of the probability
distribution given in the following table.
Outcome
-1
-1/2
0
1/2
1
Probability
1/8
3/8
1/8
1/8
2/8
Section 7.5: The Variance and Standard Deviation.
Alternative formula for variance
Alternative formula for Variance
If X is a random variable then
Var(X) = E(X 2 ) − [E(X)]2 .
Example 3
Solve the previous example using this formula.
Section 7.5: The Variance and Standard Deviation.
Standard deviation
Definition 4
σX =
p
Var(X)
Recall that we showed that if X is a binomial random variable with
parameters n and p, then E(X) = np. That proof was cute!
Variance and Standard deviation for a binomial r. v.
If X is a binomial random variable with parameters n and p, then with
q = 1 − p,
Var(X) = npq
√
σX = npq
Section 7.5: The Variance and Standard Deviation.
Example 5
Suppose that a coin is tossed 12 times. Find the mean and standard
deviation for the number of heads.
Section 7.5: The Variance and Standard Deviation.
Population Variance and Standard deviation
Definition 6
Consider a population of N numbers x1 , x2 , . . . , xN having mean µ.
Then the variance is defined as
σ2 =
(x1 − µ)2 + (x2 − µ)2 + · · · + (xN − µ)2
N
If the numbers have been grouped into the values x1 , x2 , . . . , xr with
frequencies f1 , f2 , . . . , fr then
σ2 =
1
[(x1 − µ)2 f1 + (x2 − µ)2 f2 + · · · + (xr − µ)2 fr ]
N
The standard deviation, σ , is defined as the square root of the
variance.
Section 7.5: The Variance and Standard Deviation.
Example 7
Find the variance and standard deviation for the population of scores
score
0
5
10
15
frequency
13
23
31
17
Section 7.5: The Variance and Standard Deviation.
Let’s fill out the table...
xi
0
5
10
15
TOTAL
fi
13
23
31
17
84
xi − µ
-170/21
-65/21
40/21
145/21
(xi − µ)2
(xi − µ)2 (fi )
Section 7.5: The Variance and Standard Deviation.
Sample Variance and Standard Deviation
Definition 8
Consider a sample of n numbers x1 , x2 , . . . , xn having mean x̄. Then
the variance is
s2 =
(x1 − x̄)2 + (x2 − x̄)2 + · · · + (xn − x̄)2
n−1
If the numbers have been grouped into the values x1 , x2 , . . . , xr with
frequencies f1 , f2 , . . . , fr then
s2 =
1
[(x1 − x̄)2 f1 + (x2 − x̄)2 f2 + · · · + (xr − x̄)2 fr ]
n−1
The standard deviation, s, is defined as the square root of the
variance.
Section 7.5: The Variance and Standard Deviation.
Investment returns (#5)
Example 9
The tables below gives the probability distribution for the possible
returns from two different investments.
1
Compute the mean and the variance for each investment.
2
Which investment has the highest expected return (i.e. mean)?
3
Which investment is less risky (i.e., has lesser variance)?
Investment A
Return ($ millions) Probability
-10
1/5
20
3/5
25
1/5
Investment B
Return ($ millions) Probability
0
.3
10
.4
30
.3
Section 7.5: The Variance and Standard Deviation.
We start by computing µA and µB :
Now we compute each variance σA2 and σB2 :
Section 7.5: The Variance and Standard Deviation.
Computing s sample standard deviation. In Textbook (page
340)
Compute the sample standard deviations for frequency distributions
of sales in car dealerships A and B. (Read page 311, section 7.2, on
Textbook and example 1, section 7.4, page 327).
The frequency distribution for dealership A is given in the following
table
Weekly sales Frequency
5
2
6
2
7
13
8
20
9
10
10
4
11
1
Total
n = 52
Section 7.5: The Variance and Standard Deviation.
“We consider the data from dealership A to be a sample, since it was
only one year’s data and we are interested in making comparisons and
using them to predict the future sales of the two dealerships”
You can check that x¯A = 7.96.
1
[(5 − 7.96)2 ·2+(6 − 7.96)2 ·2+(7 − 7.96)2 ·13+(8 − 7.96)2 ·
s2A = 51
20 + (9 − 7.96)2 · 10 + (10 − 7.96)2 · 4 + (11 − 7.96)2 · 1 ≈ 1.45.
So
√
sA = sA ≈ 1.20 cars.
For dealership B: sB ≈ 2.03 cars.
Since sA < sB , dealership B exhibits greater variation than dealership
A during the time that sales were observed.
Section 7.5: The Variance and Standard Deviation.
Chebychev’s inequality
Chebychev’s inequality
Suppose that a probability distribution with numerical outcomes has
expected value µ and standard deviation σ . Then the probability that
a randomly chosen outcome lies between µ − c and µ + c ,
inclusive, is at least 1 −
σ2
.
c2
Section 7.5: The Variance and Standard Deviation.
Example 10
Suppose that a probability distribution has mean 35 and standard
deviation 5. Use the Chebychev’s inequality to estimate the
probability that an outcome will lie from
(a) 25 to 45.
(b) 20 to 50.
(c) 29 to 41.