The Binomial Distribution

1

• The Binomial Distribution
• This distribution is useful for modelling
  situations in which the random variable
  (representing the outcome of an experiment) may
  take one of only two possible values.
• For example, sitting for a test, you can either
  have a success or a failure if a coin is tossed,
  the outcome is either a Head or a Tail, etc.

2

• The following additional features characterise
  the Binomial
• 1. The number of trials n is a known constant
  and it is not too large.
• 2. For each trial, the probability of success p
  is a known constant and is the same for each
  trial.

3
Binomial (25, 0.1)
4
Binomial (25, 0.9)
5
Binomial (40, 0.6) Mean np 24 S.D. (np
(1-p))1/2 3.09
6
Binomial (40, 0.6)
7
Binomial (2500, 0.001)
8
Binomial (2500, 0.001) Meannp 2.5 S.D. (np
(1-p))1/2 1.58
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• The Poisson Distribution
• Consider the following situation.
• Customers come to a shop at a regular interval
• The average rate of arrival of customers is given
  (l) but the total number of customers to arrive
  is unknown and could be very large
•  

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• The probability of getting r customers in any
  given interval is then given by
• P(X r) (e-l lr)/ r!
• where e 2.718..
• Theory
• If x Binomial (n,?), n is large, yet np is
  not large, then x has an approximate Poisson
  distribution with l n?

11
Half a percent of 500 students in a course are
likely to resort to unfair means.
Find the probability that

Exactly 4 students will resort to unfair
means0.133836 0.133602
At most 6 students will resort to unfair means
More than 4 students will resort to unfair means
Less than 4 students will resort to unfair means
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Continuous Probability Distributions
13
The Uniform distribution
This distribution is useful for modelling
situations where a random outcome may be
imagined to have realizations along a straight
line with equal probability
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Suppose that commuters waiting on a platform 50
metres long are likely to spread out evenly
Suppose we call X the location chosen by
a typical commuter.
Then 0 ? X ? 50 and X has a uniform distribution
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f(X)
For every value of X, f(X) 1/50
1/50
0 50
X
25
The Expected Value of X 25
Any realization of X is a mode
The distribution is perfectly symmetric
The Median 25
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f(X)
For every value of X, f(X) 1/(b-a)
1/(b-a)
0 a b
X
(ab)/2
The Expected Value of X (ab)/2
The distribution is perfectly symmetric
So the St. Dev. of X (b-a) /?12
Any realization of X is a mode
The variance of X (b-a)2/12
The Median (ab)/2
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f(X)
1/50
0 50
X
25
The St. Dev. of X (50-0) /?12 14.43
The variance of X (50-0)2/12 208.3
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Simulation from Uniform(0, 50)
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The Triangular distribution
Can you locate the mean median and the mode?
X
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The Triangular distribution
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Every normal distribution is symmetric about the
mean.
The two shaded parts must be equal in area.
Mean-a Mean Mean a z
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The two inner green areas are equal as well.
Mean-a Mean Mean a z
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The area shaded brown is approximately 68 of the
whole
-1 0 1
z
25
The area shaded orange is approximately90 of
the whole
z
0

• -1.645 1.645

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The area shaded orange is approximately95 of
the whole
z
-2 0 2
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f(x)
m
x
Two Normal Distribution curves with same mean
(m) but different standard deviation
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f(z)
m m2
z
Two Normal Distribution curves with same
standard deviation but different values of mean
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Find the area shaded black
Answer 0.5 0.4452 0.9452
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The area shaded black is 0.9452 as well (by
symmetry)
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Find the area shaded black
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The black shaded area below has the same area
(by symmetry)
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Answer 0.50.3944 0.8944
First find this area
This area is 0.3944
Then add 0.5
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Look for z so that this area is 0.25
The answer is 0.675
Q3 0.675
Q1 -0.675
25
25
Q1
Q3
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Look for z so thatthis area is 0.4
D9 1.28
D1 -1.28
10
10
D1
D9
Find the first decile of the SND
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So Q1 for x -0.675(5) 20 16.625
So Q3 for x 0.675(5) 20 23.375
Exercise Find the first decile of X Normal
(20, 52)
Q1 for z is 0.675
x zsm
Q3 for z is 0.675
Find the first and the third quartile of X
Normal (20, 52)
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35 of British men are at least 185 cm tall
If I meet 200 such men on any given day, what is
the probability that 100 or more of them are 185
cm or taller?
Normal Approximation of the Binomial (Chapter P4
of the text)
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Probability(M)
Prob( M 100) ?
P(99.5 lt M lt 100.5) is the normal approximation
.
.
.
.
.
.
M
99 100 101 102 103 104
39
Prob(M ? 100) Prob(M 100) Prob(M 101)
Prob(M 102) ... Prob(M 200)
This area is shaded black
But it is also the area under the red polygon to
the right of 99.5, or Prob(M gt 99.5)
Prob(M ? 100) Prob(M gt 99.5)
Probability(M)
Prob( M 100) P(99.5 lt M lt 100.5)
So
.
.
.
.
.
.
M
99 100 101 102 103 104
40
Find Prob( M lt 103)
Probability(M)
It is also the area under the red polygon or
Prob(M ? 102.5)
This area is shaded black
.
.
.
.
.
.
M
99 100 101 102 103 104
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Binomial Normal Approximation
Prob( M 100) Prob(99.5 lt M lt100.5)
Prob(M ? 100) Prob(M gt 99.5)
Prob( M lt 103) Prob(M ? 102.5)
Prob( M ? 103) Prob( M ? 103.5)
We can similarly show that
The approximation works if both of np and n(1-p)
?5
Read page 500 (inset)