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.
P ( x > 28) is adjusted to
P ( x > 28.5) = _______ x > 28
Does not include 28 It includes all values to the right of 28 28 28.5 27.5. x x > 28.5 4. z= (28.5−20) 40•.5•.5 =2.69 z z=2.69 left tail area =
.9964
right tail area = .0036
z > 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
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 so the Binomial Distribution can be considered normal 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
P ( x < 15) is adjusted to P ( x < 15.5) = _______ 15 15.5 14.5 x x < 15 includes 15
and all values to the left
x < 15.5 4. z z= (15.5−24) 1200•.02•.98 =−1.75 z =–1.75 left tail area
= .0401
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.
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 so the Binomial Distribution can be considered normal 2. µ =n• p=.09•60=54 σ = n• p•q= 60•.9•.1 3. x is at least 56 is written x > 56 P ( x > 56 ) is adjusted to P ( x > 55.5) = _______ x > 56 includes 56
and all values to the right
56 56.5 55.5 x 4. z z= (55.5−54) 60•.9•.1=.65 z =.65 left tail area
= .7422
right tail area = .2578
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.
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 so the Binomial Distribution can be considered normal 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
P ( 38 < x < 45) is adjusted to P (37.5 < x < 45.5) = _______ 38 < x < 45 includes 38 and 45
It also includes all values between 38 and 45
38 38.5 44.5 x 45 37.5 45.5 4. z z= (45.5−40) 50•.8•.2=1.94 .z=– .53
the area between z = –.53 and z = 1.94 .9738–.2981= .6757 z= (37.5−40) 50•.8•.2 =−.53 z=1.94 area left of –.53 = .2981 area left of 1.94 = .9738 5. P(37.5< x<45.5)=.6757
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.
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 so the Binomial Distribution can be considered normal 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
P ( 45 < x < 52) is adjusted to
P (45.5 < x < 51.5) = _______ 45 < x < 52
Does not include 45 to 52
It includes all values between 45 and 52
45 45.5 51.5 x 52 44.5 52.5 4. z z= (51.5−42) 60•.7•.3=2.68 .z=.99 the area between z = .99 and z = 2.68 .9963–.8389= .1574 z= (45.5−42) 60•.7•.3 =.99 z=2.68 area left of .99 = .8389 area left of 2.68 = .9963 5. P(45.5<x<51.5)=.1574
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.
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 Normal Approximation
P(between 45 and 52) (non inclusive) 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 nCx•.7x•.3n−x area between z = .99 and z = 2.68
z .z=.99
the area between z = .99 and z = 2.68 .9963–.8389= .1574 z=2.68 area left of .99 = .8389 area left of 2.68 = .9963 P(x=46)=60C46•.746•.314 =.0621447 P(x=47)=60C47•.747•.313 =.0431928 P(x=48)=60C48•.748•.312 =.0272954 P(x=49)=60C49•.749•.311=.0155974 P(x=50)=60C50•.750•.310=.0080067 P(x=51)=60C51•.751•.39 =.0036632 P(46)+P( 47)+P(48)+P(49)+P(50)+P(51) 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.
