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

2. Example: Variance and Standard Deviation of the Number of Radios Sold in a Week x, Radiosp(x), Probability(x -  X ) 2 p(x) 0p(0) = 0.03(0 – 2.1) 2 (0.03) = 0.1323 1p(1) = 0.20(1 – 2.1) 2 (0.20) = 0.2420 2p(2) = 0.50(2 – 2.1) 2 (0.50) = 0.0050 3p(3) = 0.20(3 – 2.1) 2 (0.20) = 0.1620 4p(4) = 0.05(4 – 2.1) 2 (0.05) = 0.1805 5p(5) = 0.02(5 – 2.1) 2 (0.02) = 0.1682 1.00 0.8900 Variance Standard deviation Variance and Standard Deviation µ x = 2.10

3. Expected Value and Variance (Summary) The expected value, or mean, of a random variable The expected value, or mean, of a random variable is a measure of its central location. is a measure of its central location. The expected value, or mean, of a random variable The expected value, or mean, of a random variable is a measure of its central location. is a measure of its central location. The variance summarizes the variability in the The variance summarizes the variability in the values of a random variable. values of a random variable. The variance summarizes the variability in the The variance summarizes the variability in the values of a random variable. values of a random variable. The standard deviation, , is defined as the positive The standard deviation, , is defined as the positive square root of the variance. square root of the variance. The standard deviation, , is defined as the positive The standard deviation, , is defined as the positive square root of the variance. square root of the variance.

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

5. Discrete Probability Distribution Models

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

7. Binomial Distribution Our interest is in the number of successes Our interest is in the number of successes occurring in the n trials. occurring in the n trials. Our interest is in the number of successes Our interest is in the number of successes occurring in the n trials. occurring in the n trials. We let x denote the number of successes We let x denote the number of successes occurring in the n trials. occurring in the n trials. We let x denote the number of successes We let x denote the number of successes occurring in the n trials. occurring in the n trials.

8. where: where: f ( x ) = the probability of x successes in n trials f ( x ) = the probability of x successes in n trials n = the number of trials n = the number of trials p = the probability of success on any one trial p = the probability of success on any one trial Binomial Distribution n Binomial Probability Function

9. Binomial Distribution n Binomial Probability Function Probability of a particular sequence of trial outcomes sequence of trial outcomes with x successes in n trials with x successes in n trials Probability of a particular sequence of trial outcomes sequence of trial outcomes with x successes in n trials with x successes in n trials Number of experimental outcomes providing exactly outcomes providing exactly x successes in n trials Number of experimental outcomes providing exactly 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): 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). three are unemployed. Note: (n = 15, p = 0.02). P(x= 3) = 0.0029 (from Binomial Table). P(x= 3) = 0.0029 (from Binomial Table). three or more are unemployed. three or more are unemployed. P(x  3) = 1- [0.7386 +0.2261 + 0.0323] = 0.0031. P(x  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. 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? 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. 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? 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 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 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 n 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. 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. 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 1 st Worker 2 nd Worker 3 rd Worker x x Prob. Leaves (.1) Leaves (.1) Stays (.9) Stays (.9) 3 3 2 2 0 0 2 2 2 2 Leaves (.1) S (.9) Stays (.9) S (.9) L (.1).0010.0090.7290.0090 1 1 1 1.0810 11

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

21. Poisson & Hypergeometric Distributions Optional Readings

22. End of Chapter 6