Question

The mean and variance of a binomial distribution are $6$ and $4$. The parameter $n$ is

• A. $18$
• B. $12$
• C. $10$
• D. $9$

Answer 1

The correct option is A

$18$

Explanation for the correct option.

Step 1: Find the probability of success and probability of failure.

Assume that, $p$ be the probability of success and $q$ be the probability of failure.

Therefore, $p+q=1...\left(1\right)$.

Also, the mean of a binomial distribution is given by $np$ and variance of a binomial distribution is given by $npq$.

Since, the mean of the given binomial is $4$.

Therefore, $np=6...\left(2\right)$.

Since, the variance of the given binomial is $2$.

$npq=4...\left(3\right)$

From equation $2$ and $3$.

$$\begin{gathered} 6q=4 \\ \Rightarrow q=\frac{2}{3} \end{gathered}$$

Therefore, the probability of failure is $\frac{2}{3}$.

From equation $1$.

$$\begin{gathered} p=1-\frac{2}{3} \\ \Rightarrow p=\frac{1}{3} \end{gathered}$$

Therefore, the probability of success is $\frac{1}{3}$.

Step 2: Find the parameter $n$.

Find the value of parameter $n$.

Since, the probability of success is $\frac{1}{3}$.

From equation $2$, we get

$$\begin{gathered} \frac{n}{3}=6 \\ \Rightarrow n=18 \end{gathered}$$

Therefore, the value of parameter $n$ is $18$.

Hence, option $A$ is the correct answer.