In a binomial distribution B ( n , p ) , the sum and the product of the mean and the variance are 5 and 6 respectively, then 6 ( n + p ? q ) is equal to [2023]

Question

In a binomial distribution $B (n, p)$ , the sum and the product of the mean and the variance are 5 and 6 respectively, then $6 (n + p - q)$ is equal to [2023]

Smart Achievers · Accepted Answer

(3)

Let mean of distribution = $x$

and variance of distribution = $y$

$∴$ $x + y = 5 and x y = 6$

$\Rightarrow x = 3 and y = 2$ $[∵ mean > variance]$

Now, mean of binomial distribution = $n p$

and variance of binomial distribution = $n p q$

$∴$ $x = n p and y = n p q$

$\Rightarrow n p = 3 and n p q = 2 \Rightarrow 3 \cdot q = 2 \Rightarrow q = \frac{2}{3}$

Now, we know $p + q = 1$

$\Rightarrow p = 1 - q = 1 - \frac{2}{3} = \frac{1}{3}$ and $n = \frac{3}{p} = \frac{1}{1 / 3} = 9$

$∴$ $6 (n + p - q) = 6 (9 + \frac{1}{3} - \frac{2}{3}) = 6 (\frac{27 + 1 - 2}{3}) = 52$

Solution

Q.
In a binomial distribution $B (n, p)$ , the sum and the product of the mean and the variance are 5 and 6 respectively, then $6 (n + p - q)$ is equal to [2023]

1 50

2 53

3 52

4 51

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