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Lesson 4 Probability & Normal Probability Distribution . P(A)= probability of event A = (number of outcomes of event A)/(all of the outcomes) ` a P A. 1 – P(A) = the complement of event A P(A or B) = P(A)+P(B) - P(A and B) = the union of events A and B P(A and B)= P(A) x P(B) = the intersection of independent event A and independent event B P(an impossible event) = 0 P( a certain event) = 1 P(at least one) = 1 – P(none) P(B|A)= Conditional probability of event B, given event A has an outcome = P(A and B) / P(A) A ball is drawn at random from a box containing 6 red balls, 4 white balls and 5 blue balls. Find the probability that a) it is red b) white c) blue d) not red e) red or white f)red and white a) 6 / ( 6 + 4 + 5) = 6 / 15 = 2 / 5 = 0.40 b) 4 / ( 6 + 4 + 5) = 4/ 15 = 0.267 c) 5 / (6 + 4 + 5) = 5/15 = 0.33 d) P(not red) = 1 – P(red) = 1 - 0.40 = 0.60 e) P(red union white) = (6 + 4) / (6 + 4 + 5)= .667 or 1- P(blue)= 1- .0.333 = .667 f) P(red intersect white)= P(red) x P(white)= 0.40 x 0.267 = 0.11 CONTINGENCY TABLE Wireless System Preference by Gender Male Female Type A 200 35 Type B 125 300 Type C 19 21 Total 344 356 Type A Type B Type C Total Male 0.29 0.18 0.02 0.48 Female 0.05 0.43 0.03 0.51 Total 235 425 40 700 Total 0.34 0.61 0.05 1 P(M)= P(F)= P(M) or Type A= P((F) or Type B= P(F) or Type C= P(M) and Type A= P(F) and Type C= 0.48+ 0.34-0.29 0.51+0.61-0.43 0.51+0.05-0.03 0.48 x 0.29 0.51 x 0.03 0.48 0.51 0.53 0.69 0.53 0.14 0.02 P(M)+P(A)-P(M and A) P(F)+P(B)-P(F and B) P(F)+P©-P(F and C) P(M)x P(A|M) P(F)xP(C|F) fffff Using z-scores to find Normal Probabilities: An SAT Exam had X bar= 21.0 and s= 4.7. When a nearby high school student took the exam, he scored 32. His z-score is Z= (32-21)/4.7 = 2.34 . This may also be obtained using Excel: =STANDARDIZE(32, 21, 4.7)=2.34 . Looking up the probability in a Normal Distribution table, for z=2.34 p= .9904. This means that about 1 student in 100 scored better than the nearby high school student. Assume the commute time to travel from Barnegat to Toms River is normally distributed with a mean of 65 minutes and a standard deviation of 5 minutes. a) What is the z-score for a commute time of 53 minutes? b) About what percent of commutes will be faster than 60 minutes? a) =STANDARDIZE(53, 65, 5) =z = -2.4 b) =STANDARDIZE(60, 65, 5) =z =-1.0 , approximately 16% are faster. Assume that men’s weights are normally distributed with a mean of 172 lb and a standard deviation of 29 lb. Find the probability that if an individual man is selected, his weight will be greater than 200 lb. =STANDARDIZE(200, 172, 29)= z = .9655, 83 % of men will have a weight > 200 lb. 201 lb= (201-172)=29 lb = within one standard deviation above the mean. When designing a door, how high should it be if 95% of men will fit through without bumping their head? Heights of men are normally distributed with a mean of 69.0 inches, and a standard deviation of 2.8 inches. Using Excel, NORMINV(.95, 69, 2.8)= 73.6 inches = 6.13 feet high. To find the z-scores for any value of X, use =STANDARDIZE(X, Mean, Standard deviation). To find the area to the left of a z-score in the standard normal dist., use =NORMSDIST(z). To find the area to the right of a z-score, use =1 – NORMSDIST(z). To find the area between two z-scores, when z1 < z2, use =NORMSDIST(z2) - NORMSDIST(z1). By the Empirical Rule, if X is normally distributed, approximately 68% of observations will be within 1 standard deviation of x bar. 95% within 2 standard deviations of x bar, and 99.6 % within 3 standard deviations of x bar. Age Salary zAge 22 35,000 -1.4863 23 36,000 -1.15601 24 37,000 -0.82572 25 38,000 -0.49543 26 39,000 -0.16514 27 40,000 0.165145 28 41,000 0.495434 29 42,000 0.825723 30 43,000 1.156012 31 44,000 1.486301 You see in the Excel sheet, there are 6 Ages that have z-scores between -1 and 1. 6 Ages/ 10 total Ages = 60 % Empirical Probability, compared with the Theoretical Probability for a normal distribution of 68% for z values between -1 and 1. For a large amount of data, using Excel to sort and filter data is easier, faster and more accurate. (We did not use the Salary column). TOTAL 26.5 Mean 3.02765 Standard dev. http://prod.campuscruiser.com/cruiser/occ/aaikin/NormalDistribution.xls allows you to do calculations using the Normal Distribution, knowing the mean and the standard deviation. (Right-click on the link and click “Open hyperlink.” You may view and/or save the file - it is very useful. Normal Heights & z-scores Height 62 66 61 70 72 71 73 70 68 71 69 72 66 65 68 67 69 TOTAL Mean= St. Dev.= z -1.81287 -0.64327 -2.10526 0.526316 1.111111 0.818713 1.403509 0.526316 -0.05848 0.818713 0.233918 1.111111 -0.64327 -0.93567 -0.05848 -0.35088 0.233918 12 of 17 68.2 3.42 Given 17 women’s heights, x bar and s were calculated; then z-scores were calculated using the Excel “fill handle.” Then using Filter, 12 of 17 z-scores were within the between -1 and 1 criteria which equals 70.59 % probability for this Normal distribution. (Minitab 16 graph) 70%