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A2-Level Maths:
Statistics 2
for Edexcel
Binomial distribution
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Binomial distributions
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
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Special distributions
Many real-life situations can be modelled using statistical
distributions. Examples of the types of problem that can be
addressed using these distributions include:
In a board game, players needs a six before they can start.
What is the probability that they haven’t started after 5 tries?
What proportion of the adult population have an IQ above
120?
The number of accidents on a stretch of motorway averages
1 every 2 days. How likely is it that there will be no accidents
in a week?
12% of people are left-handed. What is the probability that a
class of 30 people will have more than 6 left-handed people?
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Binomial distribution
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Binomial distribution
Introductory example:
A spinner is divided into four equal
sized sections marked 1, 2, 3, 4.
If the spinner is spun 6 times, how likely
is it to land on 1 on four occasions?
One possible sequence would be 1 1 1 1 1′ 1′.
6!
 6C4
The number of possible sequences is
4 !2 !
Most calculators
have an nCr
button
(i.e. the number of ways of arranging 6 items, where 4 are of
one kind and 2 are of a different kind).
Each sequence has probability 0.254 × 0.752.
So the required probability is 6 !  0.25 4  0.752  0.0330.
4 !2 !
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Binomial distribution
A binomial distribution arises when the following
conditions are met:
an experiment is repeated a fixed number (n) of times
(i.e., there is a fixed number of trials);
the outcomes from the trials are independent of one another;
each trial has two possible outcomes (referred to as
success and failure);
the probability of a success (p) is constant.
If the above conditions are satisfied and X is the random
variable for the number of successes, then X has a
binomial distribution. We write: X ~ B(n , p).
n and p are called
parameters.
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Binomial distribution
Which of these situations might reasonably be modelled
by a binomial distribution?
1 Joan takes a multiple choice examination
Binomial
consisting of 40 questions. X is the number of
questions answered correctly if she chooses
each answer completely at random.
2
A bag contains 6 blue and 8 green counters.
Not
James randomly picks 5 counters from the bag
binomial
without replacement. X is the number of blue
counters picked out.
Outcomes are not independent
3
A bag contains 6 blue and 8 green counters.
Jan randomly picks 5 counters from the bag,
replacing each counter before picking the next.
X is the number of blue counters picked out.
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Binomial
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Binomial distribution
Which of these situations might reasonably be
modelled by a binomial distribution?
1 Jon throws a dice repeatedly until he obtains
Not
a six. X is the number of throws he needs
binomial
before a six arises.
The number of trials is not fixed
2 Judy counts the number of silver cars
Not
that pass her along a busy stretch of road.
binomial
X is the number of silver cars that pass in
a minute.
The number of trials is not fixed
3 Josh is a mid-wife. He delivers 10 babies.
X is the number of babies that are girls.
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Binomial
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Binomial distribution
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Binomial distribution
If X ~ B(n , p), then
P( X  x)  nCx p x q n  x
for x  0,1, 2,...n
where q = 1 – p.
Example: X ~ B(12, 0.4).
Find
a) P(X = 3)
b) P(X > 1).
Number
Probability
Probability
of
ofof
x n–x
possible
sequences
successes
failures
a) P( X  3)  12C3  0.43  0.69  0.142 (to 3 s.f.)
b) P(X > 1) = 1 – P(X = 0) – P(X = 1).
P( X  0)  12C0  0.40  0.612  0.612  0.00218
P( X  1)  12C1  0.41  0.611  0.01741
So P(X > 1) = 0.980 (3 s.f.)
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Binomial distribution
Example: The probability that a baby is born a boy is 0.51.
A mid-wife delivers 10 babies. Find:
a) the probability that exactly 4 are male;
b) the probability that at least 8 are male.
a) P( X  4)  10C4  0.514  0.496  0.197
b) P( X  8)  P( X  8)  P( X  9)  P( X  10)
  10C8  0.518  0.492    10C9  0.519  0.49   0.5110
 0.04945  0.01144  0.00119
 0.0621
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Mean and variance of a binomial
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
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Mean and variance of a binomial
It can be shown that if X ~ B(n, p), then
E[X] = np
E[X] is an unbiased
estimator of the mean.
Var[X] = np(1 – p) = npq.
and
Example:
If X ~ B[16, 0.25], then
E[X] = 16 × 0.25 = 4
and
Var[X] = 16 × 025 × 0.75 = 3
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Mean and variance of a binomial
Example: If X ~ B(n, p), E(X) = 8 and Var(X) = 4.8,
calculate P(X = 5).
We can use the information provided to form 2 equations:
E[X] = np
so, np = 8
Var[X] = npq
so, npq = 4.8
Substituting the first equation into the second we find 8q = 4.8.
Therefore q = 0.6.
So, p = 0.4 and n = 8 ÷ 0.4 = 20.
Hence, X ~ B(20, 0.4).
20
5
15
So, P(X = 5) = C5  0.4  0.6  0.0746
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Use of binomial tables
Binomial distributions
Contents
Mean and variance of a binomial
Use of binomial tables
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Use of binomial tables
Tables of probabilities are available for many binomial
distributions. The tables give cumulative probabilities,
that is P(X ≤ x).
x
P(X ≤ x)
0
0.0824
1
0.3294
2
0.6471
P(X ≤ 5) = 0.9962
3
0.8740
P(X = 4) = P(X ≤ 4) – P(X ≤ 3)
= 0.9712 – 0.8740
= 0.0972
4
0.9712
5
0.9962
6
0.9998
P(X > 2) = 1 – P(X ≤ 2) = 1 – 0.6471
= 0.3529
7
1.0000
The table shows an extract for the
cumulative probabilities for a
B(10, 0.3) distribution.
We see that:
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Use of binomial tables
Example: 1 in 4 people carry a particular gene. If 20 people
are chosen at random, find the probability that:
a) exactly 3 of them carry the gene;
b) at least 6 of them carry the gene.
x
P(X ≤ x)
0
0.0032
The table shows an extract from the
cumulative probabilities for a
B(20, 0.25) distribution. We see that:
1
0.0243
2
0.0913
3
0.2252
a) P(X = 3) = P(X ≤ 3) – P(X ≤ 2)
= 0.2252 – 0.0913
= 0.1339
4
0.4148
5
0.6172
6
0.7858
b) P(X ≥ 6) = 1 – P(X ≤ 5) = 1 – 0.6172
= 0.3828
…
…
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Use of binomial tables
Examination-style question: Jan estimates that
the probability that she has to stay late at work on
any day is 0.2. She plans to keep a record over the
next 16 working days of how frequently she has to
work late. Let X denote the number of such days. x
P(X ≤ x)
0
0.0281
1
0.1407
2
0.3518
3
0.5982
4
0.7983
5
0.9184
6
0.9734
…
…
a) State an assumption needed for a binomial
distribution to be an appropriate model for X.
Assuming that a binomial distribution is
appropriate, find:
b) the probability that she stays late at least
twice;
c) the mean and the standard deviation for the
number of days she will work late.
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Use of binomial tables
a) The main assumption here would be that the event of
her staying on late at work on any particular day must
be independent of whether she had to work late on
any other day.
Note: The assumption should be
stated in the context of the question.
b) X ~ B(16, 0.2).
P(X ≥ 2) = 1 – P(X ≤ 1)
= 1 – 0.1407 = 0.8593 (from tables)
c) E[X] = np = 16 × 0.2 = 3.2
Var[X] = npq = 16 × 0.2 × 0.8 = 2.56
 s.d = 1.6
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