Binomial Theorem

The mean of a binomial distribution is $x$, and the variance is 5 .We know that the mean, $\mu = n p = x$. This means $n$ times $p$ is $x$.The formula for variance is $\sigma^2 = n p (1-p) = 5$. This means $n$ times $p$ times $(1-p)$ is 5 .From the mean formula, $p = \frac{x}{n}$.Substitute $p = \frac{x}{n}$ into the variance formula:$x \left(1 - \frac{x}{n}\right) = 5$If we open the bracket, we get: $x - \frac{x^2}{n} = 5$Take $\frac{x^2}{n}$ to the other side: $\frac{x^2}{n} = x - 5$Now, solve for $n$ : $n = \frac{x^2}{x-5}$To get a whole number for $n, x-5$ must divide $x^2$ exactly. Let's check different $x$ values:If $x=6: n = \frac{36}{1} = 3636$ is an integer.If $x=10: n = \frac{100}{5} = 2020$ is an integer.If $x=30: n = \frac{900}{25} = 3636$ is an integer.So, the possible values for $x$ are: $6,10,30$.