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KEAM 2018 Math Paper
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© examsnet.com
Question : 27
Total: 120
The mean and variance of a binomial distribution are 8 and 4 respectively. What is (X = 1) ?
1
2
8
1
2
12
1
2
6
1
2
4
1
2
5
Validate
Solution:
Let
n
and
p
be the parameters of the binomial distribution.
Mean
=
8
and variance
=
4
⇒
n
p
=
8
and
n
p
q
=
4
⇒
q
=
1
2
=
p
and
n
=
16
∴
Required probability
=
P
(
X
=
1
)
=
16
C
1
(
1
2
)
1
(
1
2
)
15
=
16
×
(
1
2
)
16
=
2
4
2
16
=
1
2
12
© examsnet.com
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