Variance and Standard Deviation The variance of a

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

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

Variance and Standard Deviation Example: Variance and Standard Deviation of the Number of Radios

Variance and Standard Deviation Example: Variance and Standard Deviation of the Number of Radios Sold in a Week x, Radios 0 1 2 3 4 5 p(x), Probability p(0) = 0. 03 p(1) = 0. 20 p(2) = 0. 50 p(3) = 0. 20 p(4) = 0. 05 p(5) = 0. 02 1. 00 µx = 2. 10 (x - X)2 p(x) (0 – 2. 1)2 (0. 03) = 0. 1323 (1 – 2. 1)2 (0. 20) = 0. 2420 (2 – 2. 1)2 (0. 50) = 0. 0050 (3 – 2. 1)2 (0. 20) = 0. 1620 (4 – 2. 1)2 (0. 05) = 0. 1805 (5 – 2. 1)2 (0. 02) = 0. 1682 0. 8900 Standard deviation Variance

Expected Value and Variance (Summary) The expected value, or mean, of a random variable

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

Expected Value and Variance (Summary) The expected value, or mean, of a random variable is a measure of its central location. E(x) = = xf (x) the variability in p The variance summarizes the values of a random variable. p p The standard deviation, is defined as the positive square root of the variance. Var(x) = 2 = (x - )2 f (x)

Discrete Probability Distribution Models

Discrete Probability Distribution Models

Binomial Distribution p 1. The experiment of consists of a sequence of n Four

Binomial Distribution p 1. The experiment of consists of a sequence of n Four Properties a Binomial Experiment 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.

Binomial Distribution Our interest is in the number of successes occurring in the n

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.

Binomial Distribution n Binomial Probability Function where: f (x) = the probability of x

Binomial Distribution n 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

Binomial Distribution n Binomial Probability Function Number of experimental outcomes providing exactly x successes

Binomial Distribution n Binomial Probability Function Number of experimental outcomes providing exactly x successes in n trials Probability of a particular sequence of trial outcomes with x successes in n trials

Thinking Challenge Example You’re a telemarketer selling service contracts for Macy’s. You’ve sold 20

Thinking Challenge Example 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 Solutions A. P(0) =. 0687 B. P(2) =. 2835 C. P(at most

Thinking Challenge Solutions 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(12) = = P(2) + P(3). . . + 1 - [P(0) + P(1)] 1 -. 0687 -. 2062. 7251

Thinking Challenge. Example n The Department of Labor Statistics for the state of n

Thinking Challenge. Example n The Department of Labor Statistics for the state of n n 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.

Another Example n A city engineer claims that 50% of the bridges in the

Another Example n 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. n What is the probability that exactly 6 of the bridges need repair? This situation meets the binomial requirements. Why? n VERIFY. n = 10, p = 0. 5, P(x = 6) = 0. 2051. Use Binomial Table

Example Continued n What is the probability that 7 or fewer of the bridges

Example Continued n What is the probability that 7 or fewer of the bridges need repair? n We need P(x £ 7) = P(x = 0) + P(x = 1) +. . . + P(x = 7) = 0. 001 + 0. 0098 +. . . + 0. 1172 = 0. 9454 n OR P(x £ 7) = 1 – P(x=8) – P(x=9) – P(x=10) = 1 – (. 0439+. 0098+. 0010) = 0. 9454 Use Binomial Table

Binomial Distribution n More Example: Evans Electronics Wendy is concerned about a low retention

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.

Binomial Distribution Example (Continued) Choosing 3 hourly employees at random, what is the probability

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

Binomial Distribution p Tree Diagram 1 st Worker 2 nd Worker Leaves (. 1)

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

Binomial Distribution n Expected Value (Mean) E(x) = = np n Variance Var(x) =

Binomial Distribution n Expected Value (Mean) E(x) = = np n Variance Var(x) = 2 = np (1 - p ) n Standard Deviation

Binomial Distribution: Example (Continued) p p Evans is concerned about a low retention rate

Binomial Distribution: Example (Continued) p p 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 n Expected Value (Mean) E(x) = = 3(. 1) =. 3 employees

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

Poisson & Hypergeometric Distributions Optional Readings

Poisson & Hypergeometric Distributions Optional Readings

End of Chapter 6

End of Chapter 6