2. Your 3rd quiz will cover sections 4.1-4.3

3. a){HHH,HTT,THT,TTH,THH,HTH,HHT,TTT} {0,1,2,3} b) {1/8,3/8,3/8,1/8} d) P(x=2 or x=3)= P(x=2)+P(x=3)=3/8+1/8=1/2

4. Thinking Challenge A couple plans to have children until they get a girl, but they agree that they will not have more than three children even if they are all boys. (Assume boys and girls are equally likely) Create a probability model for the number of children they will have.

5. Thinking Challenge solution Sample space={G,BG,BBG,BBB} X= number of children x123 P(x)1/21/42/8

6. 1.Expected Value (Mean of probability distribution) Weighted average of all possible values  = E(x) =  x p(x) 2.Variance Weighted average of squared deviation about mean  2 = E[(x    (x    p(x)=  x 2 p(x)-  2 Summary Measures 3. Standard Deviation

7. Summary Measures Calculation Table xp(x)p(x)x p(x)x –  Total  (x    p(x) (x –   (x –   p(x) xp(x)xp(x)

8. Thinking Challenge Sample space={G,BG,BBG,BBB} X= number of children Find the expected number of children E(X)=1*1/2+2*1/4+3*2/8=7/4=1.75 Find the standard deviation of children they will have  2 = 1*1/2+4*1/4+9*2/8-(7/4) 2 =.6875  = (0.6875) 1/2 =0.83 x123 P(x)1/21/42/8

9. Thinking Challenge Sample space={G,BG,BBG,BBB} X= number of boys they will have Find the expected number of boys they will have E(X)=1*1/4+2*1/8+3*1/8=7/8=0.875 Find the standard deviation of children they will have  2 = 1*1/4+4*1/8+9*1/8-(7/8) 2 = 1.109  = (1.109) 1/2 =1.053 x0123 P(x)1/21/41/8

10. Thinking Challenge You toss 2 coins. You’re interested in the number of tails. What are the expected value, variance, and standard deviation of this random variable, number of tails?

11. Expected Value & Variance Solution* 0.25–1.001.00 1.5000 2.251.00 0.50  =1.0 xp(x)p(x)x p(x)x –  (x –   (x –   p(x).25 0    

12. Probability Rules for Discrete Random Variables Let x be a discrete random variable with probability distribution p(x), mean µ, and standard deviation . Then, depending on the shape of p(x), the following probability statements can be made: Chebshev's Rule Empirical Rule P(  -  ≤x≤  +  )≥00.68 P(  -2  ≤x≤  +2  )≥3/40.95 P(  -3  ≤x≤  +3  )≥8/40.997

13. a) 34.5, 174.75,13.219 c) 34.5 ±2*13.219=8.062,60.938 The probability that x will fall in this interval is 1.00. Under the empirical rule we expect that this value would be 0.95.

14. 4.3 The Binomial Distribution

15. Binomial Distribution Number of ‘successes’ in a sample of n observations (trials) Number of reds in 15 spins of roulette wheel Number of defective items in a batch of 5 items Number correct on a 33 question exam Number of customers who purchase out of 100 customers who enter store (each customer is equally likely to purchase)

16. Binomial Probability Characteristics of a Binomial Experiment 1.The experiment consists of n identical trials. 2.There are only two possible outcomes on each trial. We will denote one outcome by S (for success) and the other by F (for failure). 3.The probability of S remains the same from trial to trial. This probability is denoted by p, and the probability of F is denoted by q. Note that q = 1 – p. 4.The trials are independent. 5.The binomial random variable x is the number of S’s in n trials.

17. Binomial Probability Distribution Example Experiment: Toss 1 coin 5 times in a row. Note number of tails. What’s the probability of 3 tails? © 1984-1994 T/Maker Co.

18. Binomial Probability Distribution p(x) = Probability of x ‘Successes’ p=Probability of a ‘Success’ on a single trial q=1 – p n=Number of trials x=Number of ‘Successes’ in n trials (x = 0, 1, 2,..., n) n – x=Number of failures in n trials

19. Binomial Probability Table (Portion) n = 5p k.01…0.50….99 0.951….031….000 1.999….188….000 21.000….500….000 31.000….812….001 41.000….969….049 Cumulative Probabilities P(x=3)=P(x ≤ 3) – P(x ≤ 2) =.812 –.500 =.312

20. Binomial Distribution Characteristics n = 5 p = 0.1 n = 5 p = 0.5 Mean Standard Deviation

21. Thinking Challenge You’re taking a 33 question multiple choice test. Each question has 4 choices. Clueless on 1 question, you decide to guess. What’s the chance you’ll get it right? If you guessed on all 33 questions, what would be your grade? Would you pass? X= number of correct answers Binomial random variable since we just guess the answer. Total number of trials=33 p= 0.25 Expected number of correct answers=33*0.25=8.25

22. Binomial Distribution Thinking Challenge 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?

23. Binomial Distribution Solution* n = 12, p =.20 E(X)=n*p=12*0.2=2.4  =(np(1-p)) 1/2 =(12*0.2*0.8) 1/2 =1.38 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

24. Using binomial table n = 10, p =.20 A. P(X = 0) = p(0) = P(X ≤ 0) =.107 (Table I)

25. Using binomial table n = 10, p =.20 B.P(X = 2) = p(2) = P(X ≤ 2) – P(X ≤ 1) =.678 -.376 =.302

26. Using binomial table n = 10, p =.20 C. P(at most 2) = P(X ≤ 2) =.678

27. Using binomial table n = 10, p =.20 D.P(at least 2) = P(X ≥ 2) = 1 – P(X ≤ 1) = 1 –.376 =.724

28. By using TI-84: B. P(X = 2) = p(2) = P(X ≤ 2) – P(X ≤ 1) binomcdf(10,.20,2) - binomcdf(10,.20,1)

29. a) Adult not working during summer vacation. b) The experiment consist of 10 identical trials. A trial for this experiment is an individual. There are only two possible outcomes: work or do not work The probability remains same for each individual (trial) Individuals are independent c) 0.35 d) 0.2522 e) 0.2616

30. Thinking Challenge The communications monitoring company Postini has reported that 91% of e-mail messages are spam. Suppose your inbox contains 25 messages. What are the mean and standard deviation of the number of real messages you should expect to find in your inbox? What is the probability that you will find only 1 or 2 real messages?

31. Thinking Challenge The communications monitoring company Postini has reported that 91% of e-mail messages are spam. Suppose your inbox contains 25 messages. What are the mean and standard deviaiton of the number of real messages you should expect to find in your inbox? (2.25, 1.43) What is the probability that you will find only 1 or 2 real messages? (0.5117)