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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.
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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
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Expected Value and Variance (Summary)
The expected value, or mean, of a random variable is a measure of its central location. 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 variance summarizes the variability in the values of a random variable.
The standard deviation, , is defined as the positive square root of the variance. The standard deviation, , is defined as the positive square root of the variance.
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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)Var(x) = 2 = (x - )2f(x)
E(x) = = xf(x)E(x) = = xf(x)
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DiscreteProbabilityDistribution
BinomialHyper-
GeometricNegativeBinomial
Poisson
Discrete Probability Distribution ModelsDiscrete Probability Distribution Models
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Binomial Distribution
Four Properties of a Binomial Experiment
3. The probability of a success, denoted by p, does not change from trial to trial.3. The probability of a success, denoted by p, does not change from trial to trial.
4. The trials are independent.4. The trials are independent.
2. Two outcomes, success and failure, are possible on each trial.2. Two outcomes, success and failure, are possible on each trial.
1. The experiment consists of a sequence of n identical trials.1. The experiment consists of a sequence of n identical trials.
stationarity
assumption
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Binomial Distribution
Our interest is in the number of successes occurring in the n trials. 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. We let x denote the number of successes occurring in the n trials.
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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 xn
f x p px n x
Binomial Distribution
Binomial Probability Function
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( )!( ) (1 )
!( )!x n xn
f x p px 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
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
Number of experimental outcomes providing exactly
x successes in n trials
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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 ExampleThinking Challenge Example
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A. P(0) = .0687
B. P(2) = .2835
C. 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 SolutionsThinking Challenge Solutions
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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 ExampleThinking Challenge Example
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Another ExampleAnother 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 Use Binomial Table
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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 ContinuedExample Continued
Use Binomial Table Use Binomial Table
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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.
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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 xn
x n xp px n x( )
!!( )!
( )( )
1
1 23!(1) (0.1) (0.9) 3(.1)(.81) .243
1!(3 1)!f
Let: p = .10, n = 3, x = 1
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Tree DiagramBinomial Distribution
1st Worker 1st Worker 2nd Worker2nd Worker 3rd Worker3rd Worker xx Prob.Prob.
Leaves (.1)Leaves (.1)
Stays (.9)Stays (.9)
33
22
00
22
22
Leaves (.1)Leaves (.1)
Leaves (.1)Leaves (.1)
S (.9)S (.9)
Stays (.9)Stays (.9)
Stays (.9)Stays (.9)
S (.9)S (.9)
S (.9)S (.9)
S (.9)S (.9)
L (.1)L (.1)
L (.1)L (.1)
L (.1)L (.1)
L (.1)L (.1) .0010.0010
.0090.0090
.0090.0090
.7290.7290
.0090.0090
11
11
.0810.0810
.0810.0810
.0810.0810
11
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Binomial Distribution
(1 )np p
E(x) = = np
Var(x) = 2 = np(1 - p)
Expected Value (Mean)
Variance
Standard Deviation
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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)
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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
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Poisson & Hypergeometric Distributions
Optional Readings
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End of Chapter 6