Binominal Distributions Worksheet With Answer Key - Helm 2008 Section 372 Page 12
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20Answer
The table of values appropriate to this case is:
2
2
x
x
P(X = x)
xP(X = x)
x
P(X = x)
3
0
0
q
0
0
2
2
2
1
1
3q
p
3q
p
3q
p
2
2
2
2
4
3qp
6qp
12qp
3
3
3
3
9
p
3p
9p
2
2
3
2
Hence
E(X) =
xP(X = x) = 0 + 3q
p + 6qp
+ 3p
= 3p(q + p)
= 3p
since q + p = 1
2
2
V (X) = E(X
)
[E(X)]
2
2
3
2
= 0 + 3q
p + 12qp
+ 9p
(3p)
2
2
= 3p(q
+ 4qp + 3p
3p)
2
2
= 3p((1
p)
+ 4(1
p)p + 3p
3p)
2
2
2
= 3p(1
2p + p
+ 4p
4p
+ 3p
3p) = 3p(1
p) = 3pq
From the results given above, it is reasonable to asert the following result in Key Point 5.
Key Point 5
Expectation and Variance of the Binomial Distribution
If a random variable X which can assume the values 0, 1, 2, 3, . . . , n follows a binomial distribution
X
B(n, p) so that
P(X = r) =
C p q
=
C p (1
p)
then the expectation and variance of the distribution are given by the formulae
E(X) = np
and
V (X) = np(1
p) = npq
Task
A die is thrown repeatedly 36 times in all. Find E(X) and V (X) where X is the
number of sixes obtained.
Your solution
28
HELM (2008):
Workbook 37: Discrete Probability Distributions
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