Chapter 6 8B: Examples of Using a Normal Distribution to Approximate a Binomial Probability Distribution Example 1 The probability of having a boy in any single birth is 50%. Use a normal distribution to approximate the probability of getting more than 28 boys in 40 births. Is it unusual to get more than 28 boys in 40 births? Given: p =.5 n = 40 q = 1 p =.5 1. n p =.5 40 = 20 n q =.5 40 = 20 n p 5 and n q 5 so the Binomial Distrubtion can be considered normal 2. µ = n p =.5 40 = 20 σ = n p q = 40.5.5 3. x is more than 28 is written x > 28. 28 27.5. 28.5 x x > 28 Does not include 28 It includes all values to the right of 28 x > 28.5 P ( x > 28) P ( x > 28.5) = (28.5 20) 40.5.5 = 2.69 left tail area =.9964 right tail area =.0036 =2.69 > 2 so it would be unusual to have more than 28 boys in 40 births. 5. P(x > 28.5) =.0036 The probability of having more than 28 boys in 40 births is.0036 The lay person would say that there is a.4% chance of having more than 28 boys in 40 births. Stat 300 6 6B Lecture Page 1 of 6 2012 Eitel
Example 2 Laser eye surgery is becoming very common. 1200 people in Sacramento had the procedure in the last 12 months. The probability of having a serious vision problem after laser surgery is 2%. Use a normal distribution to approximate the probability that no more than 15 patients out of the 1200 have a serious vision problem. Given: p =.02 q = 1 p =.98 and n = 1200 1. n p =.02 1200 = 24 n q =.98 1200 =1176 n p 5 and n q 5 2. µ = n p =.02 1200 = 24 σ = n p q = 1200.02.98 3. x is no more than 15 is written x < 15 x < 15 includes 15 and all values to the left P ( x < 15) 15 15 15.5 x x < 15.5 P ( x < 15.5) = (15.5 24) 1200.02.98 = 1.75 left tail area =.0401 1.75 5. P(x < 15.5) =.0401 The probability that no more than 15 patients out of the 1200 have a serious vision problem is.0401. The lay person may if say that if 1200 people have laser eye surgery then there is about a 4% chance that no more than 15 of the patients will have a serious vision problem. Stat 300 6 6B Lecture Page 2 of 6 2012 Eitel
Example 3 The probability that a drug cures a patient is 90%. Use a normal distribution to approximate the probability that at least 56 out of 60 people who take the drug are cured. Round P(x) to 2 decimal places Given: p =.9 q = 1 p =.1 and n = 60 1. n p =.9 60 = 54 n q =.1 60 = 6 n p 5 and n q 5 2. µ = n p =.09 60 = 54 σ = n p q = 60.9.1 3. x is at least 56 is written x > 56 x > 56 includes 56 and all values to the right P ( x > 56 ) 56 55.5 56.5 x P ( x > 55.5) = (55.5 54) 60.9.1 =.65 left tail area =.7422 right tail area =.2578.65 5. P(x > 55.5) =.2578 The probability that at least 56 patients out of the 60 are cured by the drug is.0.2578 The lay person might if say that if 60 people use the drug then there is about a 26% chance that at least 56 of the patients will be cured. Stat 300 6 6B Lecture Page 3 of 6 2012 Eitel
Example 4 The probability that a plane lands on time at the Sacramento Airport is 80%. Use a normal distribution to approximate the probability that between 38 and 45 ( inclusive ) out of 50 people who take the drug are cured. Round P(x) to 2 decimal places Given: p =.7 q = 1 p =.3 and n = 60 1. n p =.8 50 = 40 n q =.2 50 = 10 n p 5 and n q 5 2. µ = n p =.8 50 = 40 σ = n p q = 50.8.2 3. x is between 38 and 45 ( inclusive ) is written 38 < x < 45 38 < x < 45 includes 38 and 45 It also includes all values between 38 and 45 P ( 38 < x < 45) 38 45 37.5 38.5 45 45.5 x P (37.5 < x < 45.5) = area left of 1.94 =.9738 area left of.53 =.2981 (37.5 40) 50.8.2 =.53 (45.5 40) 50.8.2 = 1.94 5. P(37.5 < x < 45.5) =.6757.=.53 the area between.53 and 1.94.9738.2981=.6757 =1.94 The probability that between 38 and 45 (inclusive) out of 50 planes land on time at the Sacramento Airport is.6757 A lay person might if say that if 50 planes land at the Sacramento Airport there is about a 66% chance that between 38 and 45 ( inclusive ) out of 50 land on time. Stat 300 6 6B Lecture Page 4 of 6 2012 Eitel
Example 5 The probability that a drug cures a patient is 70%. Use a normal distribution to approximate the probability that between 45 and 52 ( non inclusive ) out of 60 people who take the drug are cured. Round P(x) to 2 decimal places Given: p =.7 q = 1 p =.3 and n = 60 1. n p =.7 60 = 42 n q =.3 60 =18 n p 5 and n q 5 2. µ = n p =.7 60 = 42 σ = n p q = 60.7.3 3. x is between 45 and 52 ( non inclusive ) is written 45 < x < 52 45 < x < 52 Does not include 45 to 52 It includes all values between 45 and 52 P ( 45 < x < 52) 45 52 45 45.5 51.5 52.5 x P (45.5 < x < 51.5) = area left of 2.68 =.9963 area left of.99 =.8389 (45.5 42) 60.7.3 =.99 (51.5 42) 60.7.3 = 2.68 5. P(45.5 < x < 51.5) =.1574.=.99 the area between.99 and 2.68.9963.8389=.1574 =2.68 The probability that between 45 and 52 ( non inclusive) out of 60 people who take the drug are cured is.1574 A lay person might say that if 60 people take the drug then there is about a 16% chance that between 45 and 52 (non inclusive) are cured. Stat 300 6 6B Lecture Page 5 of 6 2012 Eitel
What is the difference between using the Binomial Calculation versus using the Normal Approximation? Lets do Example 5 both ways and see The probability that a drug cures a patient is 70%. Use a normal distribution to estimate the probability that between 45 and 52 ( non inclusive) out of 60 people who take the drug are cured. Given: p =.7 n = 60 q = 1 p =.3 Binomial Technique P(between 45 and 52) (non inclusive) Normal Approximation P(between 45 and 52) using the normal curve = P(46) + P(47) + P(48) + P(49) + P(50) + P(51) with continuity corrections P(45.5 < x < 51.5) Binomial Probabilities calculated P(45.5 < x < 51.5) using n C x.7 x.3 n x area between.99 and 2.68 P(x = 46)= 60 C 46.7 46.3 14 =.0621447 P(x = 47)= 60 C 47.7 47.3 13 =.0431928 area left of 2.68 =.9963 area left of.99 =.8389 P(x = 48)= 60 C 48.7 48.3 12 =.0272954 P(x = 49)= 60 C 49.7 49.3 11 =.0155974 P(x = 50)= 60 C 50.7 50.3 10 =.0080067 P(x = 51)= 60 C 51.7 51.3 9 =.0036632 P(46) + P(47) + P(48) + P(49) + P(50) + P(51).=.99 the area between.99 and 2.68.9963.8389=.1574 =2.68 P(between 45 and 52) =.1599 P(45.5 < x < 51.5) =.1574 The Normal approximation is off by.0025 (25 ten thousandths). This is with only 6 rectangles. The normal approximation gets closer and closer as the number of rectangles increases. Stat 300 6 6B Lecture Page 6 of 6 2012 Eitel