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Binomial Random Variables Binomial Probability Distributions Binomial Random Variables Through 2/25/2014 NC State’s free-throw percentage is 65.1% (315th out 351 in Div. 1). If in the 2/26/2014 game with UNC, NCSU shoots 11 free-throws, what is the probability that: NCSU makes exactly 8 free-throws? NCSU makes at most 8 free throws? NCSU makes at least 8 free-throws? “2-outcome” situations are very common Heads/tails Democrat/Republican Male/Female Win/Loss Success/Failure Defective/Nondefective Probability Model for this Common Situation Common characteristics ◦ repeated “trials” ◦ 2 outcomes on each trial Leads to Binomial Experiment Binomial Experiments n identical trials ◦ n specified in advance 2 outcomes on each trial ◦ usually referred to as “success” and “failure” p “success” probability; q=1-p “failure” probability; remain constant from trial to trial trials are independent Classic binomial experiment: tossing a coin a pre-specified number of times Toss a coin 10 times Result of each toss: head or tail (designate one of the outcomes as a success, the other as a failure; makes no difference) P(head) and P(tail) are the same on each toss trials are independent ◦ if you obtained 9 heads in a row, P(head) and P(tail) on toss 10 are same as P(head) and P(tail) on any other toss (not due for a tail on toss 10) Binomial Random Variable The binomial random variable X is the number of “successes” in the n trials Notation: X has a B(n, p) distribution, where n is the number of trials and p is the success probability on each trial. Examples a. b. c. d. Yes; n=10; success=“major repairs within 3 months”; p=.05 No; n not specified in advance No; p changes Yes; n=1500; success=“chip is defective”; p=.10 Binomial Probability Distribution n trials, p success probability on each tri al probability distribution: p(x) n C x p q x nx , x 0,1, 2, n E ( x) x0 n xp ( x ) ,n xx p q n x0 V ar ( x ) E ( x npq x nx np Rationale for the Binomial Probability Formula P(x) = n! • (n – x )!x! Number of outcomes with exactly x successes among n trials px • n-x q Binomial Probability Formula P(x) = n! • (n – x )!x! Number of outcomes with exactly x successes among n trials px • n-x q Probability of x successes among n trials for any one particular order Graph of p(x); x binomial n=10 p=.5; p(0)+p(1)+ … +p(10)=1 The sum of all the areas is 1 Think of p(x) as the area of rectangle above x p(5)=.246 is the area of the rectangle above 5 Binomial Probability Histogram: n=100, p=.5 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 70 68 66 64 62 60 58 56 54 52 50 48 46 44 42 40 38 36 34 32 30 0 Binomial Probability Histogram: n=100, p=.95 0.18 0.17 0.16 0.15 0.14 0.13 0.12 0.11 0.1 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 Example A production line produces motor housings, 5% of which have cosmetic defects. A quality control manager randomly selects 4 housings from the production line. Let x=the number of housings that have a cosmetic defect. Tabulate the probability distribution for x. Solution (i) D=defective, G=good outcome x P(outcome) GGGG 0 (.95)(.95)(.95)(.95) DGGG 1 (.05)(.95)(.95)(.95) GDGG 1 (.95)(.05)(.95)(.95) : : : DDDD 4 (.05)4 Solution ( ii ) x is a binom ial random variable p(x) n C x p q x nx , x 0,1, 2, ,n n 4, p .05 ( q .95) p (0) 4 C 0 (.05) (.95) .815 0 4 p (1) 4 C 1 (.05) (.95) .171475 1 3 p (2) 4 C 2 (.05) (.95) .01354 2 2 p (3) 4 C 3 (.05) (.95) .00048 3 1 p (4) 4 C 4 (.05) (.9 5) .00000625 4 0 Solution x 0 p(x) .815 1 2 .171475 .01354 3 4 .00048 .00000625 Example (cont.) x 0 p(x) .815 1 2 .171475 .01354 3 .00048 4 .00000625 What is the probability that at least 2 of the housings will have a cosmetic defect? P(x p(2)+p(3)+p(4)=.01402625 Example (cont.) x 0 p(x) .815 1 2 .171475 .01354 3 4 .00048 .00000625 What is the probability that at most 1 housing will not have a cosmetic defect? (at most 1 failure=at least 3 successes) P(x 3)=p(3) + p(4) = .00048+.00000625 = .00048625 Using binomial tables; n=20, p=.3 9, 10, 11, … , 20 P(x 5) = .416 P(x > 8) = 1- P(x 8)= 1- .887=.113 =P(x 8) P(x < 9) = ? 8, 7, 6, … , 0 P(x 10) = ? 1- P(x 9) = 1- .952 P(3 x 7)=P(x 7) - P(x 2) .772 - .035 = .737 Binomial n = 20, p = .3 (cont.) P(2 < x 9) = P(x 9) - P(x 2) = .952 - .035 = .917 P(x = 8) = P(x 8) - P(x 7) = .887 - .772 = .115 Color blindness The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population. We can model this situation with a B(n = 25, p = 0.08) distribution. What is the probability that five individuals or fewer in the sample are color blind? Use Excel’s “=BINOMDIST(number_s,trials,probability_s,cumulative)” P(x ≤ 5) = BINOMDIST(5, 25, .08, 1) = 0.9877 What is the probability that more than five will be color blind? P(x > 5) = 1 P(x ≤ 5) =1 0.9877 = 0.0123 What is the probability that exactly five will be color blind? P(x = 5) = BINOMDIST(5, 25, .08, 0) = 0.0329 x P (X = x) P (X < = x) 0 1 2 1 2 .4 4 % 2 7 .0 4 % 2 8 .2 1 % 1 2 .4 4 % 3 9 .4 7 % 6 7 .6 8 % 30% 3 4 1 8 .8 1 % 9 .0 0 % 8 6 .4 9 % 9 5 .4 9 % 25% 5 6 7 8 3 .2 9 % 0 .9 5 % 0 .2 3 % 0 .0 4 % 9 8 .7 7 % 9 9 .7 2 % 9 9 .9 5 % 9 9 .9 9 % 9 10 11 0 .0 1 % 0 .0 0 % 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % 12 13 0 .0 0 % 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % 5% 14 15 0 .0 0 % 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % 0% 16 17 18 0 .0 0 % 0 .0 0 % 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % 19 20 0 .0 0 % 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % 21 22 0 .0 0 % 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % Probability distribution and histogram for 23 24 0 .0 0 % 0 .0 0 % 1 0 0 .0 0 % 1 0 0 .0 0 % the number of color blind individuals among 25 0 .0 0 % 1 0 0 .0 0 % B(n = 25, p = 0.08) 15% 24 22 20 18 16 14 12 10 8 6 4 2 10% 0 P(X = x) 20% Number of color blind individuals (x) 25 Caucasian males. What are the expected value and standard deviation of the count X of color blind individuals in the SRS of 25 Caucasian American males? E(X) = np = 25*0.08 = 2 SD(X) = √np(1 p) = √(25*0.08*0.92) = 1.36 What if we take an SRS of size 10? Of size 75? E(X) = 10*0.08 = 0.8 E(X) = 75*0.08 = 6 SD(X) = (75*0.08*0.92)=2.35 0.5 0.2 0.4 0.15 0.3 p = .08 n = 10 0.2 0.1 P(X=x) P(X=x) SD(X) = √(10*0.08*0.92) = 0.86 p = .08 n = 75 0.1 0.05 0 0 0 1 2 3 4 5 Number of successes 6 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Number of successes Recall Free-throw question Through 2/25/14 NC State’s free-throw percentage was 65.1% (315th in Div. 1). If in the 2/26/14 game with UNC, NCSU shoots 11 freethrows, what is the probability that: 1. 2. 3. NCSU makes exactly 8 free-throws? NCSU makes at most 8 free throws? NCSU makes at least 8 free-throws? 1. n=11; X=# of made free-throws; p=.651 p(8)= 11C8 (.651)8(.349)3 =.226 2. P(x ≤ 8)=.798 3. P(x ≥ 8)=1-P(x ≤7) =1-.5717 = .4283 Recall from beginning of Lecture Unit 4: Hardee’s vs The Colonel Out of 100 taste-testers, 63 preferred Hardee’s fried chicken, 37 preferred KFC Evidence that Hardee’s is better? A landslide? What if there is no difference in the chicken? (p=1/2, flip a fair coin) Is 63 heads out of 100 tosses that unusual? Use binomial rv to analyze n=100 taste testers x=# who prefer Hardees chicken p=probability a taste tester chooses Hardees If p=.5, P(x 63) = .0061 (since the probability is so small, p is probably NOT .5; p is probably greater than .5, that is, Hardee’s chicken is probably better). Recall: Mothers Identify Newborns After spending 1 hour with their newborns, blindfolded and nose-covered mothers were asked to choose their child from 3 sleeping babies by feeling the backs of the babies’ hands 22 of 32 women (69%) selected their own newborn “far better than 33% one would expect…” Is it possible the mothers are guessing? Can we quantify “far better”? Use binomial rv to analyze n=32 mothers x=# who correctly identify their own baby p= probability a mother chooses her own baby If p=.33, P(x 22)=.000044 (since the probability is so small, p is probably NOT .33; p is probably greater than .33, that is, mothers are probably not guessing.