Download 6-8B Normal Distribution to Approximate a Binomial Probability

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 • p ≥ 5 and n • q ≥ 5
2. µ = n • p = .5 • 40 = 20
n • q = .5• 40 = 20
so the Binomial Distrubtion can be considered normal
σ = n • p• q = 40 • .5 • .5
3. x is more than 28 is written x > 28.
x > 28
Does not include 28
It includes all values to the
right of 28
P ( x > 28)
is adjusted to
28
27.5.
P ( x > 28.5) = _______
28.5
x
x > 28.5
4.
z=
(28.5 − 20)
= 2.69
40• .5 • .5
left tail area =
.9964
right tail
area =
.0036
z=2.69
z
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
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
1. n • p = .02•1200 = 24
and n = 1200
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
x < 15
includes 15
and all values to the left
P ( x < 15)
is adjusted to
15
14.5
15.5
x < 15.5
P ( x < 15.5) = _______
x
4.
(15.5 − 24)
z=
= −1.75
1200 • .02• .98
left tail area
= .0401
z =–1.75
5.
z
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
1. n • p = .9• 60 = 54
n • p ≥ 5 and n • q ≥ 5
2. µ = n • p = .09 • 60 = 54
and n = 60
n • q = .1• 60 = 6
so the Binomial Distribution can be considered normal
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 )
is adjusted to
56
55.5
P ( x > 55.5) = _______
56.5
x
4.
z=
(55.5 − 54)
= .65
60• .9• .1
left tail area
= .7422
right tail area
= .2578
z =.65
5.
z
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
1. n • p = .8• 50 = 40
n • p ≥ 5 and n • q ≥ 5
and
n = 60
n • q = .2• 50 = 10
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
38 < x < 45
includes 38 and 45
It also includes all values between 38 and 45
P ( 38 < x < 45)
is adjusted to
38
37.5
45
44.5
38.5
P (37.5 < x < 45.5) = _______
45.5
x
4.
area left of 1.94 = .9738
area left of –.53 = .2981
z=
(37.5 − 40)
= −.53
50• .8• .2
z=
(45.5 − 40)
= 1.94
50• .8• .2
.z=– .53
z=1.94 z
the area between
z = –.53 and z = 1.94
.9738–.2981= .6757
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.
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
1. n • p = .7• 60 = 42
n • p ≥ 5 and n • q ≥ 5
and
n = 60
n • q = .3• 60 = 18
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
45 < x < 52
Does not include 45 to 52
It includes all values between 45 and 52
P ( 45 < x < 52)
is adjusted to
45
44.5
52
51.5
45.5
P (45.5 < x < 51.5) = _______
52.5
x
4.
area left of 2.68 = .9963
area left of .99 = .8389
z=
(45.5 − 42)
= .99
60 • .7• .3
z=
(51.5 − 42)
= 2.68
60 • .7• .3
.z=.99
z=2.68 z
the area between
z = .99 and z = 2.68
.9963–.8389= .1574
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.
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
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
nC x
• .7 x • .3n− x
area between z = .99 and z = 2.68
area left of 2.68 = .9963
P( x = 46)= 60 C46 • .746 • .314 = .0621447
P( x = 47)= 60 C47 • .747 • .313 = .0431928
area left of .99 = .8389
P( x = 48)= 60 C48 • .748 • .312 = .0272954
P( x = 49)= 60 C49 • .749 • .311 = .0155974
P( x = 50)= 60 C 50 • .7 50 • .310 = .0080067
51
9
P( x = 51)= 60 C 51 • .7 • .3
.z=.99
= .0036632
z=2.68 z
the area between
z = .99 and z = 2.68
.9963–.8389= .1574
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.
Stat 300 6 – 6B Lecture
Page 6 of 6
© 2012 Eitel