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Normal Approximation to Binomial
Lesson 5: Continuous Random Variables
§5.3 Normal Approximation to Binomial
Satya Mandal, KU
September 20
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
Preview
Normal Approximation to Binomial
”As far as the laws of mathematics refer to reality, they are
not certain; and as far as they are certain, they do not refer to
reality”. - Albert Einstein
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Goals
◮
Give a method to approximate Binomial random
varibales, by normal.
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
Normal Approximation
◮
Bionomial random variables X ∼ B(n, p) was discussed in
last chapter. Two methods to compute probability were
given
◮
By hand, using the formula for the probability function:
p(r ) = P(X = r ) =n Cr p r (1 − p)r
◮
◮
r = 0, 1, 2, . . . , n
The formula was not used in this class.
To compute probability binomialcdf function was used.
It was commented before, most random varibles in nature
and life are Normal or can be approximated by Normal
random variables. In this section, we discuss how to use
Normal to approximate X ∼ B(n, p).
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
Review the Animation for the Bell Shape
◮
◮
For some visual justification for using normal
approximation for X ∼ B(n, p) review Animation 5.3.1.
The graph of the probability function y = p(x) has the
following properties:
◮
◮
◮
◮
◮
◮
◮
It maintains a bell-shape - perfect or not.
It peaks at (or near) the mean x = µ = np.
It has a perfect Bell-shape when p = .5.
It is skewed to the left, if p < .5
It is skewed to the right, if p > .5
If the number of trials n is large, it is more like a bell.
So, we expect good approximation, by normal, when
p = .5 and n is large, else this may not be as good.
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
A Glitch
◮
A X ∼ B(n, p), random variable is discrete and
P(X = r ) =n Cr p r (1 − p)r 6= 0
◮
r = 0, 1, 2, . . . , n
A Y ∼ N(µ, σ) is continuous. If attempt to use Y ,
naively, to approximate X , then
P(X = r ) ≈ P(Y = r ) = P(r ≤ Y ≤ r ) = 0
◮
The Remedy: To remedy this glitch, we approximate
P(X = r ) ≈ P(r − .5 ≤ Y ≤ r + .5)
◮
r = 0, 1, 2, . . . , n
The following theorem states all that fully.
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
The Approximation Theorem
Theorem. Suppose X is a binomial random variable
(X ∼ B(n, p)). Assume the number of trials n is large and p is
not too close to 0 and 1. Then,
◮ Then X behaves, approximately, like a normal random
variable (Y ∼ N(µ, σ)), with mean µ and standard
deviation σ, given by
p
µ = np, and
σ = np(1 − p).
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
Continued
◮
More precisely, for integers 0 ≤ r ≤ s,
P(r ≤ X ≤ s) = P(r − .5 ≤ X ≤ s + .5)
r − .5 − µ
s + .5 − µ
≈P
≤Z ≤
σ
σ
r − .5 − µ s + .5 − µ
,
= normalcdf
σ
σ
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
Remarks
Remarks
◮ (1) The adjustment by ±.5 above, is called the continuity
correction. This was needed because we are
approximating a discrete random variable, by a
continuous random variable. However, if n is large it will
not make any detectable difference.
◮ Note, the binomialpdf from of TI-84 gives exact
probability, while normal approxiamtion is an
approximation. In any case, TI-84 will give up if n is large.
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
Exercise 5.3.3.
The campaign committee of a candidate claims that sixty
percent of the voters are in favor of the candidate. You
interview 150 voters. Assuming that the campaign committe’s
claim is accurate, what is the approximate probability that less
than 77 will favor the candidate? Solution:
◮ Here p = .6 and n = 150. First step is to compute the
mean µ and the standard deviation σ:
◮
µ = np = 150 ∗ .6 = 90, and
p
p
σ = np(1 − p) = 150 ∗ .6 ∗ (1 − .6) = 6
Let X = number of voters in favor of the candidate.
Then X ∼ B(150, .6).So, approximately, X ∼ N(90, 6).
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
Continued
◮
Now, ”X is less than 77” means ”X < 77”, which is
”X ≤ 76”. So,
P(X ≤ 76) = P(X ≤ 76.5)
[This is continuity correction.]
76.5 − µ
76.5 − µ
X −µ
≤
≈P
=P Z ≤
σ
σ
σ
76.5 − 90
) = P(Z < −2.25)
6
= normalcdf (−5, −2.25) = .0122
= P(Z <
◮
Alternately, using TI-84, exact probability
= binomialcdf (150, .6, 76) = .01278 Note that they are
real close.
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
Exercise 5.3.4.
A technique is used to fertilize eggs in a fertility clinic
laboratory. It is known that the probability that an egg will be
fertilized by this technique is 0.1. If 500 eggs are treated, what
is the probability that at least 60 eggs will be fertilized?
Solution:
◮
◮
Here p = .1 and n = 500. First step is to compute the
mean µ and the standard deviation σ:
µ = np = 500 ∗ .1 = 50
and
p
p
σ = np(1 − p) = 500 ∗ .1 ∗ (1 − .1) = 6.7082
Let X = number of eggs that will be fertilize. Then
X ∼ B(500, .1). So, approximately, X ∼ N(50, 6.7082).
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
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Normal Approximation to Binomial
Problems
Continued
◮
Now ”X is at least 60” means ”60 ≤ X ”.
P(60 ≤ X ) = P(59.5 ≤ X )
[This is continuity correction.]
59.5 − µ
59.5 − µ
X −µ
=P
=P
≤
≤Z
σ
σ
σ
59.5 − 50
≤ Z = P(1.4162 ≤ Z )
=P
6.7082
= normalcdf (1.4162, 5) = .0784
◮
Alternately, using TI-84, exact probability
= 1 − binomialcdf (500, .1, 59) = .0810 They differ too
much, becaue p = .1 is too close to 0.
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
Preview
Normal Approximation to Binomial
Problems
Exercise 5.3.10.
Suppose that an insurance company knows from experience
that the probability that a life-insurance policyholder will
survive another 10 years is p = 0.9. The company has 2280
policy holders. What is the probability that more than 2025
will survive another 10 years. Solution:
◮ Here p = .9 and n = 2280. First step is to compute the
mean µ and the standard deviation σ:
◮
µ = np = 2280 ∗ .9 = 2052
and
p
p
σ = np(1 − p) = 2280 ∗ .9 ∗ (1 − .9) = 14.3248
Let X = number of policy holders who survive another 10
years. Then X ∼ B(2280, .9). So, approximately,
X ∼ N(2052, 14.3248).
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim
Preview
Normal Approximation to Binomial
Problems
Continued
◮
Now ”more than 2025” means ”2025 < X ” which is
”2026 ≤ X ”.
P(2026 ≤ X ) = P(2025.5 ≤ X )
[This is continuity correctio
2025.5 − µ
X −µ
2025.5 − µ
=P
≤
≤Z
=P
σ
σ
σ
2025.5 − 2052
≤ Z = P(−1.8499 ≤ Z )
=P
14.3248
= normalcdf (−1.8499, 5) = .9678
◮
Alternately, using TI-84, exact probability
= 1 − binomialcdf (2280, .9, 2025) = .9663 Fairly close!
My TI-83 gave up, becaue n is too large.
Satya Mandal, KU
Lesson 5: Continuous Random Variables §5.3 Normal Approxim