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PubH 6414 Worksheet 6b: Binomial and Poisson Distributions 1 of 6 Example 1: Probability Distribution for a Discrete random variable Number of people in a family in the US (2005 data). Number of family members Probability 2 0.42 3 0.228 4 0.209 5 0.092 6 0.032 7+ 0.018 Use the probability distribution table above to calculate the following probabilities for the number of people in US families in 2005. Let X represent the number of people in a family. a. P(X ≥ 4) = P(X=4) + P(X=5) + P(X=6) + P(X=7 or more) = .209 + .092 + .032 + .018 = 0.351 b. P(3 ≤ X ≤ 5) = P(X=3) + P(X=4) + P(X=5 = .228 + .209 +.092 = .529 c. P(X < 4) = P(X=2) + P(X=3) = 0.42 + 0.228 = 0.648 Example 2: Binomial Distribution 1. Number of heads in 2 coin tosses i. What are the 4 possible outcomes from 2 coin tosses: HH TT HT TH ii. Probability of 0, 1, 2 heads in 2 coin tosses calculated from the possible outcomes Number of heads Probability 0 1/4 = 0.25 1 2/4 = 0.5 2 1/4 = 0.25 iii. Probability of 0, 1, 2 heads calculated from the binomial distribution formula n = 2, π = 0.5 2! 2 *1 0.50 (1 − 0.5) 2 − 0 = * 1 * 0.25 = 1 * 0.25 = 0.25 P(X = 0) = 0!(2 − 0)! 1 * 2 *1 2! 2 *1 0.51 (1 − 0.5) 2 −1 = * 0.5 * 0.5 = 2 * 0.25 = 0.5 P(X = 1) = 1!(2 − 1)! 1 *1 2! 2 *1 0.52 (1 − 0.5) 2 − 2 = * 0.25 * 1 = 1 * 0.25 = 0.25 P(X = 2) = 2!( 2 − 2)! 2 * 1 *1 PubH 6414 Worksheet 6b: Binomial and Poisson Distributions 2 of 6 2. Probability Distribution for X~ B (10,0.08) Number of patients surviving 5 years 0 1 2 3 4 5 6 7 8 9 10 Probability < 0.0001 <0.0001 0.0001 0.0008 0.0055 0.0264 0.0881 0.2013 0.3020 0.2684 0.1074 Use the probability distribution above for X ~ B(10, 0.8) to calculate the probability that the indicated number of patients survive at least 5 years. i. P(X = 8) = 0.3020 ii. P(X > 8) = P(X=9) + P(X=10) = 0.2684 + 0.1074 = 0.3758 iii. P(X < 7) = 1 - P(X ≥ 7) = 1 – [ P (X=7) + P(X=8) + P(X > 8)] = 1 – [0.2013 + 0.3020 + 0.3758] = 1 - 0.8791 = 0.1209 PubH 6414 Worksheet 6b: Binomial and Poisson Distributions 3. Probability distribution for X ~ B(8, 0.03) Number of deaths from heart attack 0 Probability 1 0.194 2 0.021 3 0.0013 4 < 0.001 5 <0.001 6 <0.001 7 <0.001 8 <0.001 3 of 6 0.784 Use the probability distribution above for X ~ B(8, 0.03) to calculate the probability that the indicated number of patients die from a heart attack i. P(X > 1) = 1 – P(X ≤ 1) = 1 – [P(X=0) + P(X=1)] = 1-[0.784 + 0.194] = 1-0.978 = 0.022 ii. P( X < 2) = P(X=0) + P(X=1) = 0.784 +0.194 = 0.978 iii. P(1 ≤ X ≤ 2) = P(X=1) + P(X=2) = 0.194 +0.021 = 0.215 iv. P(all 8 patients survive) = P(X = 0) = 0.784 PubH 6414 Worksheet 6b: Binomial and Poisson Distributions 4 of 6 Example 3: Normal approximation to the binomial distribution The risk of osteoporosis being diagnosed in the next year for a group of women age 70+ with osteopenia is 0.50. a. For a group of 100 women in this population identify the parameters of the binomial distribution where ‘success’ = diagnosis of osteoporosis in the next year. n = 100 π = 0.50 mean of binomial distribution = 100*0.50 = 50 b. Is the normal approximation to the binomial distribution appropriate? nπ = 50 n(1-π) =50 Both are > 5 so normal approximation to the binomial is appropriate. c. What are the mean and standard deviation for this normal approximation? mean = n*π = 50 variance = n*π(1−π) = 50*0.5 = 25 std. deviation = square root of variance = 5 d. How would you calculate the probability that between 40 and 60 women in this population will be diagnosed with osteoporosis in the next year? Draw the normal curve approximation, the area that corresponds to this probability. P(40-60 women diagnosed) = 0.9772499 - 0.02275013 = 0.9544997 = 0.9545 Draw a normal distribution curve with mean at 50 and SD = 5. Notice most of the curve is contained between 35 and 65. Draw lines at 40 and 60, which is the area between 2 SD below the mean and 2 SD above the mean. PubH 6414 Worksheet 6b: Binomial and Poisson Distributions 5 of 6 Use Rcmdr to get the probability. If you use the menu option, you must subtract the probability below 40 from that below 60. Rcmdr Menu: Distributions > Continuous Distributions > Normal Distribution > Normal Probabilities Variable Values = 40, 60; mu=50; sigma=5; select Lower Tail R Script: pnorm(60, 50, 5)-pnorm(40, 50, 5) Example 4: Normal approximation to the Poisson distribution At a large urban hospital, the average number of emergency department admissions on a Saturday night is 120. On one Saturday in July, 160 patients were admitted to the emergency department. Use the normal approximation to the Poisson distribution to determine the probability of admitting 160 or more patients given the expected number of admissions. a. Is the normal approximation appropriate? Yes, the average rate / Saturday = 120 which is > 10. b. What is the mean of the normal approximation? 120 c. What is the standard deviation of the normal approximation? sqrt(120) = 10.95 d. Draw the normal approximation curve and the area that represents > 160 patients admitted. Draw a normal distribution curve with mean at 120 and SD = 10.95. Most of the area will be between 87.2 and 152.9. Since 160 is beyond 3 SD above the mean, the probability of admitting 160 patients given the typical admission of 120 is very small. PubH 6414 Worksheet 6b: Binomial and Poisson Distributions 6 of 6 e. Calculate this probability. P(admitting > 160) = 0.0001296133 = 0.00013 Rcmdr Menu: Distributions > Continuous Distributions > Normal Distribution > Normal Probabilities Variable Value = 160; mu=120; sigma=10.95; select Upper Tail R Script: pnorm(160, 120, 10.95, lower.tail=FALSE) 1-pnorm(160, 120, 10.95)