Variance and Standard Deviation The variance of a discrete random variable is: The standard...

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 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.

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)

DiscreteProbabilityDistribution

BinomialHyper-

GeometricNegativeBinomial

Poisson

Discrete Probability Distribution ModelsDiscrete 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.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


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.

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 Probability Function

( )!( ) (1 )

!( )!x n xn

f x p px n x


!!( )!

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

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

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

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

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

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


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.


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

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



S (.9)S (.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


(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)


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

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