Download Normal approximation

A2-Level Maths:
Statistics 2
for Edexcel
S2.3 Normal
approximation to
Binomial
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Approximating the binomial using a normal
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Approximating the binomial using a normal
Calculating probabilities using the binomial distribution can
be cumbersome if the number of trials (n) is large.
Consider this example:
Introductory example:10% of people in the United
Kingdom are left-handed.
A school has 1 200 students. Find the probability that
more than 140 of them are left-handed.
Let the number of left-handed people in the school be X.
Then X ~ B[1200, 0.1].
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Approximating the binomial using a normal
The required probability is
P(X > 140) = P(X = 141) + P(X = 142) + … + P(X = 1200).
As no tables exist for this distribution, calculating this
probability by hand would be a mammoth task.
A further problem arises if you attempt to work out one of
these probabilities, for example P(X = 141):
P( X  141)  1200 C141  0.1141  0.91059
Calculators cannot calculate
the value of this coefficient –
it is too large!
One way forward is to approximate the binomial
distribution using a normal distribution.
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Approximating the binomial using a normal
Key result: If X ~ B(n, p) where n is large and p is small, then
X can be reasonably approximated using a normal distribution:
X ≈ N[np, npq]
where q = 1 – p.
There is a widely used rule of thumb that can be applied
to tell you when the approximation will be reasonable:
A binomial distribution can be approximated
reasonably well by a normal distribution
provided np > 5 and nq > 5.
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Approximating the binomial using a normal
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Approximating the binomial using a normal
A continuity correction must be applied when approximating
a discrete distribution (such as the binomial) to a continuous
distribution (such as the normal distribution).
Continuity correction:
Exact distribution: B(n, p)
P(X ≥ x)
Approximate distribution:
N[np, npq]
P(X ≥ x – 0.5)
This 0.5 is called the
continuity correction
factor.
P(X ≤ x)
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P(X ≤ x + 0.5)
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Approximating the binomial using a normal
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Approximating the binomial using a normal
Introductory example (continued): 10% of people in
the United Kingdom are left-handed.
A school has 1 200 students. Find the probability that
more than 140 of them are left-handed.
Solution:
Let the number of left-handed people in the school be X.
Then the exact distribution for X is X ~ B[1200, 0.1].
Since np = 120 > 5 and nq = 1080 > 5 we can approximate
this distribution using a normal distribution:
X ≈ N[120, 108].
np
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npq
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Approximating the binomial using a normal
So, P(X > 140) = P(X ≥ 141) → P(X ≥ 140.5)
Using continuity
correction
N[120, 108]
Standardize
N[0, 1]
140.5  120
 1.973
108
You convert 140.5 to the
standard normal distribution
using the formula:
Z
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X 

~ N[0,1].
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Approximating the binomial using a normal
Therefore P(X ≥ 140.5) = P(Z ≥ 1.973)
= 1 – Φ(1.973)
= 1 – 0.9758
= 0.0242
So the probability of there being more than 140
left-handed students at the school is 0.0242.
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Approximating the binomial using a normal
Example: It has been estimated that 15% of
schoolchildren are short-sighted. Find the probability
that in a group of 80 schoolchildren there will be
a) no more than 15 children that are short-sighted
b) exactly 10 children that are short-sighted.
Solution:
Let the number of short-sighted children in the group be X.
Then the exact distribution for X is X ~ B[80, 0.15].
Since np = 12 > 5 and nq = 68 > 5 we can approximate this
distribution using a normal distribution:
X ≈ N[12, 10.2].
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Approximating the binomial using a normal
a) So P(X ≤ 15) → P(X ≤ 15.5)
N[12, 10.2]
Using continuity correction
Standardize
N[0, 1]
15.5  12
 1.096
10.2
Therefore P(X ≤ 15.5) = P(Z ≤ 1.096)
= Φ(1.096)
= 0.8635
So the probability that no more than 15
children will be short-sighted is 0.8635.
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Approximating the binomial using a normal
b) So P(X = 10) → P(9.5 ≤ X ≤ 10.5)
Using continuity correction
Standardize
9.5  12
 0.783
10.2
N[12, 10.2]
N[0, 1]
10.5  12
 0.470
10.2
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Approximating the binomial using a normal
Therefore P(9.5 ≤ X ≤ 10.5) = P(–0.783 ≤ Z ≤ –0.470)
= P(0.470 ≤ Z ≤ 0.783)
= 0.7832 – 0.6808 = 0.1024
The probability that 10 children will be short-sighted is 0.1024.
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Examination-style question
Examination-style question:
A sweet manufacturer makes sweets in 5 colours. 25% of
the sweets it produces are red.
The company sells its sweets in tubes and in bags. There
are 10 sweets in a tube and 28 sweets in a bag. It can be
assumed that the sweets are of random colours.
a) Find the probability that there are more than 4 red
sweets in a tube.
b) Using a suitable approximation, find the probability
that a bag of sweets contains between 5 and 12 red
sweets (inclusive).
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Examination-style question
Solution:
a) Let the number of red sweets in a tube be X.
Then the exact distribution for X is X ~ B[10, 0.25].
This distribution cannot be approximated by a normal but its
probabilities are tabulated:
P(X > 4) = 1 – P(X ≤ 4)
= 1 – 0.9219
= 0.0781
So the probability that a tube contains more than 4 red
sweets is 0.0781.
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Examination style question
Solution:
b) Let the number of red sweets in a bag be Y.
Then the exact distribution for Y is Y ~ B[28, 0.25].
This distribution can be approximated by a normal since
np = 7 and nq = 21 (both greater than 5):
Y ≈ N[7, 5.25]
P(5 ≤ Y ≤ 12) → P(4.5 ≤ Y ≤ 12.5)
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npq
Using continuity correction
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Examination style question
Standardize
N[7, 5.25]
4.5  7
 1.091
5.25
12.5  7
 2.400
5.25
N[0, 1]
Therefore P(4.5 ≤ Y ≤ 12.5) = P(–1.091 ≤ Z ≤ 2.400)
= P(Z ≤ 2.400) – P(Z ≤ –1.091)
= Φ(2.400) – (1 – Φ(1.091))
= 0.9918 – (1 – 0.8623) = 0.8541
So the probability that a bag will contain between
5 and 12 red sweets is 0.8541.
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