Question

In a Binomial distribution, the mean is $15$ and variance is $10$. Then, parameter $n$ is

• A. $28$
• B. $16$
• C. $45$
• D. $25$

Answer 1

The correct option is C

$45$

Explanation for the correct option.

Find the value of the parameter $n$.

In a binomial distribution, $n$ is the number of trials, $p$ is the probability of success, and $q$ is the probability of failure.

The mean is given by the term $np$. It is given that mean is $15$, so $np=15...\left(1\right)$.

The variance is given by the term $npq$. It is given that the variance is $10$, so $npq=10...\left(2\right)$.

Divide equation $2$ by equation $1$.

$$\begin{gathered} \frac{npq}{np}=\frac{10}{15} \\ \Rightarrow q=\frac{2}{3} \end{gathered}$$

It is known that $p+q=1$, so the value of $p$ is:

$$\begin{gathered} p+q=1 \\ \Rightarrow p+\frac{2}{3}=1\left[q=\frac{2}{3}\right] \\ \Rightarrow p=1-\frac{2}{3} \\ \Rightarrow p=\frac{1}{3} \end{gathered}$$

Now find the value of $n$ by substituting $\frac{1}{3}$ for $p$ in the equation $1$.

$$\begin{gathered} np=15 \\ \Rightarrow n\times \frac{1}{3}=15 \\ \Rightarrow n=15\times 3 \\ \Rightarrow n=45 \end{gathered}$$

So the value of parameter $n$ is $45$.

Hence, the correct option is C.