“Teach A Level Maths”
Statistics 1
The Binomial Distribution:
Mean and Variance
© Christine Crisp
The Binomial Distribution – Mean and Variance
Statistics 1
AQA
MEI/OCR
OCR
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The Binomial Distribution – Mean and Variance
There are very simple formulae that we can use to
find the mean and variance of a Binomial Distribution.
I’ll illustrate the formulae with an example.
The Binomial Distribution – Mean and Variance
Suppose X ~ B(3, 0  4) . The probability distribution is
P ( X  0)  0  216
Can you find the mean using
the formula for any discrete
probability distribution?
P ( X  1)  0  432
P ( X  2)  0  288
P ( X  3)  0  064
If you’ve forgotten how to calculate the mean, the results
written in a table may remind you:
x
0
1
2
3
P ( X  x ) 0  216 0  432 0  288 0  064
mean, m =
 xP( X  x)  0  0  216  1 0  432 
...
 1 2
Can you spot the link to the parameters of the distribution
( n = 3 and p = 0·4 )?
The Binomial Distribution – Mean and Variance
Suppose X ~ B(3, 0  4) . The probability distribution is
P ( X  0)  0  216
Can you find the mean using
the formula for any discrete
probability distribution?
P ( X  1)  0  432
P ( X  2)  0  288
P ( X  3)  0  064
If you’ve forgotten how to calculate the mean, the results
written in a table may remind you:
x
0
1
2
3
P ( X  x ) 0  216 0  432 0  288 0  064
mean, m =
 xP( X  x)  0  0  216  1 0  432 
 1 2
ANS: 1  2  3  0  4 so, m  np
...
The Binomial Distribution – Mean and Variance
X ~ B(3, 0  4)
Now for the variance:
x
0
1
2
3
P ( X  x ) 0  216 0  432 0  288 0  064
The formula for the variance of any discrete probability
distribution is
 x P( X  x)  m
2
2
 0 2  0  216  1 2  0  432 
2
2
2  0  288  3 2  0  064  1 2
 0  72
The link with the parameters isn’t so easy to spot this
time but it’s easy to remember. It is
variance  npq
The Binomial Distribution – Mean and Variance
e.g.1 Find the mean and variance of the random
variable X where
X ~ B(10, 0  3)
Solution:
mean, m  np
 m  10  0  3  3
variance  npq

variance  10  0  3  0  7  2  1
SUMMARY
The mean of a Binomial Distribution is given by
mean, m  np
The variance is given by
variance  npq
The Binomial Distribution – Mean and Variance
Exercise
1. Find the mean and variance of the random variable X
where
(a) X ~ B(20, 0  5)
(b) X ~ B(15, 0  3)
(c) X ~ B(7, 0  02)
(d) X ~ B(12, 0  25)
Solutions:
(a) n  20, p  0  5, q  0  5  mean  np  10
variance  npq  5
(b) n  15, p  0  3, q  0  7  mean  np  4 5
variance  npq  3 15
The Binomial Distribution – Mean and Variance
Exercise
1. Find the mean and variance of the random variable X
where
(a) X ~ B(20, 0  5)
(b) X ~ B(15, 0  3)
(c) X ~ B(7, 0  02)
(d) X ~ B(12, 0  25)
Solutions:
(c) n  7, p  0  02, q  0  98  mean  np  0  14
variance  npq  0  1372
(d) n  12, p  0  25, q  0  75  mean  np  3
variance  npq  2  25
The following slide contains repeats of
information on earlier slides, shown without
colour, so that it can be printed and
photocopied.
The Binomial Distribution – Mean and Variance
SUMMARY
The mean of a Binomial Distribution is given by
mean, m  np
The variance is given by
variance  npq
e.g.1 Find the mean and variance of the random
variable X where
X ~ B(10, 0  3)
Solution:
mean, m  np  m  10  0  3  3
variance  npq

variance  10  0  3  0  7  2  1