Variance and Standard Deviation

1. Variance and Standard Deviation • The variance of a discrete random variable is: • The standard deviation is the square root of the variance.

2. Variance and Standard Deviation Standard deviation Variance Example: Variance and Standard Deviation of the Number of Radios Sold in a Week x, Radiosp(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 µx = 2.10

3. 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.

4. E(x) =  = xf(x) Var(x) =  2 = (x - )2f(x) 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.

5. Discrete Probability Distribution Models

6. Binomial Distribution Four Properties of a Binomial Experiment • 1. The experiment consists of a sequence of n • identical trials. • 2. Two outcomes, success and failure, are possible • on each trial. 3. The probability of a success, denoted by p, does not change from trial to trial. stationarity assumption 4. The trials are independent.

7. 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.

8. Binomial Distribution • Binomial Probability Function 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

9. Binomial Distribution • 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

10. 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 of A. No sales? B. Exactly 2 sales? C. At most 2 sales? D. At least 2 sales? Thinking Challenge Example

11. 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 Solutions

12. 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

13. 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 exactly6 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

14. 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

15. 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.

16. 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. Let: p = .10, n = 3, x = 1

17. Tree Diagram Binomial Distribution x 1st Worker 2nd Worker 3rd Worker Prob. L (.1) .0010 3 Leaves (.1) .0090 2 S (.9) Leaves (.1) L (.1) .0090 2 Stays (.9) 1 .0810 S (.9) L (.1) 2 .0090 Leaves (.1) Stays (.9) 1 .0810 S (.9) L (.1) 1 .0810 Stays (.9) 0 .7290 S (.9)

18. Binomial Distribution E(x) =  = np • Var(x) =  2 = np(1 -p) • Expected Value (Mean) • Variance • Standard Deviation

19. 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)

20. Binomial Distribution E(x) =  = 3(.1) = .3 employees out of 3 Var(x) =  2 = 3(.1)(.9) = .27 • Expected Value (Mean) • Variance • Standard Deviation

21. Poisson & Hypergeometric Distributions Optional Readings

22. End of Chapter 6