The mean and variance of a binomial distribut | vedclass

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(A) For a binomial distribution with parameters $n$ and $p$,the mean is $\mu = np = x$ and the variance is $\sigma^2 = npq = 5$,where $q = 1 - p$.Sinc

Vedclass · Accepted Answer

$6, 10, 30$

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(A) For a binomial distribution with parameters $n$ and $p$,the mean is $\mu = np = x$ and the variance is $\sigma^2 = npq = 5$,where $q = 1 - p$.Since $np = x$ and $npq = 5$,we have $xq = 5$,which implies $q = \frac{5}{x}$.Since $0  5$.Also,$p = 1 - q = 1 - \frac{5}{x} = \frac{x-5}{x}$.Since $0  5$.We know that $n = \frac{x}{p} = \frac{x}{(x-5)/x} = \frac{x^2}{x-5}$.Since $n$ must be a positive integer,$x-5$ must be a divisor of $x^2$.We can write $x^2 = (x-5)(x+5) + 25$,so $n = x+5 + \frac{25}{x-5}$.For $n$ to be an integer,$x-5$ must be a divisor of $25$.The divisors of $25$ are $1, 5, 25$.Case $1$: $x-5 = 1 \implies x = 6$. Then $n = 6+5 + 25/1 = 36$.Case $2$: $x-5 = 5 \implies x = 10$. Then $n = 10+5 + 25/5 = 20$.Case $3$: $x-5 = 25 \implies x = 30$. Then $n = 30+5 + 25/25 = 36$.Thus,the possible values for $x$ are $6, 10, 30$.

The mean and variance of a binomial distribution are $x$ and $5$ respectively. If $x$ is an integer,then the possible values for $x$ are

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