Download - Variance and Standard Deviation
Variance and Standard Deviation The variance of a discrete random
variable is:
xAll
XX xpx )()( 22
2XX
The standard deviation is the square root of the variance.
Example: Variance and Standard Deviation of the Number of Radios Sold in a Week
x, Radios p(x), Probability (x - X)2 p(x) 0 p(0) = 0.03 (0 – 2.1)2 (0.03) =
0.1323 1 p(1) = 0.20 (1 – 2.1)2 (0.20) =
0.2420 2 p(2) = 0.50 (2 – 2.1)2 (0.50) =
0.0050 3 p(3) = 0.20 (3 – 2.1)2 (0.20) =
0.1620 4 p(4) = 0.05 (4 – 2.1)2 (0.05) =
0.1805 5 p(5) = 0.02 (5 – 2.1)2 (0.02) =
0.1682 1.00
0.8900
89.02 X
Variance
9434.089.0 X
Standard deviation
Variance and Standard Deviation
µx = 2.10
Expected Value and Variance (Summary)
The expected value, or mean, of a random variable is a measure of its central location.
The variance summarizes the variability in the values of a random variable.
The standard deviation, , is defined as the positive square root of the variance.
Expected Value and Variance (Summary) The expected value, or mean, of a random
variable is a measure of its central location.
The variance summarizes the variability in the values of a random variable.
The standard deviation, is defined as the positive square root of the variance.
Var(x) = 2 = (x - )2f(x)
E(x) = = xf(x)
DiscreteProbabilityDistribution
Binomial Hyper-Geometric
NegativeBinomial Poisson
Discrete Probability Distribution Models
Binomial Distribution Four Properties of a Binomial Experiment
3. The probability of a success, denoted by p, does not change from trial to trial.
4. The trials are independent.
2. Two outcomes, success and failure, are possible on each trial.
1. The experiment consists of a sequence of n identical trials.
stationarity
assumption
Binomial Distribution
Our interest is in the number of successes occurring in the n trials.
We let x denote the number of successes occurring in the n trials.
where: f(x) = the probability of x successes in n trials n = the number of trials p = the probability of success on any one trial
( )!( ) (1 )!( )!x n xnf x p p
x n x
Binomial Distribution
Binomial Probability Function
( )!( ) (1 )!( )!x n xnf x p p
x n x
Binomial Distribution
!!( )!
nx n x
( )(1 )x n xp p
Binomial Probability Function
Probability of a particular sequence of trial outcomes with x successes in n trials
Number of experimental outcomes providing exactly
x successes in n trials
You’re a telemarketer selling service contracts for Macy’s. You’ve sold 20 in your last 100 calls (p = .20). If you call 12 people tonight, what’s the probability ofA. No sales?B. Exactly 2 sales?C. At most 2 sales? D. At least 2 sales?
Thinking Challenge Example
A. P(0) = .0687 B. P(2) = .2835C. P(at most 2) = P(0) + P(1) + P(2)
= .0687 + .2062 + .2835= .5584
D. P(at least 2) = P(2) + P(3)...+ P(12)
= 1 - [P(0) + P(1)] = 1 - .0687 - .2062= .7251
Thinking Challenge Solutions
The Department of Labor Statistics for the state of Kentucky reports that 2% of the workforce in Treble County is unemployed. A sample of 15 workers is obtained from the county. Compute the following probabilities (Hint - Binomial):
three are unemployed. Note: (n = 15, p = 0.02). P(x= 3) = 0.0029 (from Binomial Table). three or more are unemployed. P(x ³ 3) = 1- [0.7386 +0.2261 + 0.0323] = 0.0031.
Thinking Challenge Example
Another Example A city engineer claims that 50% of the bridges
in the county needs repair. A sample of 10 bridges in the county was selected at random.
What is the probability that exactly 6 of the bridges need repair? This situation meets the binomial requirements. Why?
VERIFY. n = 10, p = 0.5, P(x = 6) = 0.2051.
Use Binomial Table
What is the probability that 7 or fewer of the bridges need repair?
We need P(x £ 7) = P(x = 0) + P(x = 1) + ... + P(x = 7) = 0.001 + 0.0098 + ... + 0.1172 = 0.9454
OR P(x £ 7) = 1 – P(x=8) – P(x=9) – P(x=10) = 1 – (.0439+.0098+.0010) = 0.9454
Example Continued
Use Binomial Table
Binomial Distribution
More Example: Evans Electronics Wendy is concerned about a low retention rate for employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.
Binomial Distribution
Example (Continued) Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? Useing the equation.
f x nx n x
p px n x( ) !!( )!
( )( )
1
1 23!(1) (0.1) (0.9) 3(.1)(.81) .2431!(3 1)!f
Let: p = .10, n = 3, x = 1
Tree DiagramBinomial Distribution
1st Worker 2nd Worker 3rd Worker x Prob.
Leaves (.1)
Stays (.9)
3
2
0
22
Leaves (.1)
Leaves (.1)S (.9)
Stays (.9)
Stays (.9)
S (.9)
S (.9)
S (.9)
L (.1)
L (.1)
L (.1)
L (.1) .0010
.0090
.0090
.7290
.0090
11
.0810
.0810
.0810
1
Binomial Distribution
(1 )np p
E(x) = = np
Var(x) = 2 = np(1 p)
Expected Value (Mean)
Variance
Standard Deviation
Evans is concerned about a low retention rate for employees. In recent years, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employee chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year.
Choosing 3 hourly employees at random, what is the probability that 1 of them will leave the company this year? What is the mean, variance and the standard deviation?
Binomial Distribution: Example (Continued)
Binomial Distribution
3(.1)(.9) .52 employees
E(x) = = 3(.1) = .3 employees out of 3
Var(x) = 2 = 3(.1)(.9) = .27
Expected Value (Mean)
Variance
Standard Deviation
Poisson & Hypergeometric Distributions
Optional Readings
End of Chapter 6