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प्रश्न
The difference between mean and variance of a binomial distribution is 1 and the difference of their squares is 11. Find the distribution.
उत्तर
np - npq = 1 ...(1)
⇒ np (1 - q) = 1
⇒ npp = 1
⇒ np2 = 1
Also n2p2 - n2p2q2 = 11
⇒ n2p2(1 - q2) = 11
⇒ n2p2(1 - q) ( 1 + q) = 11
⇒ n2p2 . p . (1 + q) = 11
⇒ np2 np (1 + q) = 11
⇒ np ( 1 + q) = 11
⇒ np - npq = 1 ...(1)
`((np + npq = 11) ...(2))/( 2np = 12)`
np = 6
and np2 = 1
⇒ 6p = 1
⇒ p = `(1)/(6)`
∴ q = 1 - p = 1 - `(1)/(6) = (5)/(6)`
Now, np = 6
⇒ `n(1/6) = 6`
⇒ n = 36
So, the binomial distribution is :
(q + p)n i.e., `(5/6 + 1/6)^36`
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