Question

In binomial probability distribution, the mean is $3$ and the standard deviation is $\frac{3}{2}.$ Then the probability distribution is:

• A. ${\left(\frac{3}{4}+\frac{1}{4}\right)}^{12}$
• B. ${\left(\frac{1}{4}+\frac{3}{4}\right)}^{12}$
• C. ${\left(\frac{1}{4}+\frac{3}{4}\right)}^9$
• D. ${\left(\frac{3}{4}+\frac{1}{4}\right)}^9$

Answer 1

The correct option is A

${\left(\frac{3}{4}+\frac{1}{4}\right)}^{12}$

Explanation for the correct option:

Step-1 : Finding the probability:

Let $p$ be the probability of success and $q$ be the probability of failure for the given binomial probability distribution.

Then, the probability distribution will be ${\left(p+q\right)}^n$, where $n$ is the number of trials and $p+q=1$.

We know that, for a binomial distribution with the parameters $n,p,q$, the mean and the standard deviation will be $m=np$ and $\sigma =\sqrt{npq}$.

Given that, the mean is $3$ and the standard deviation is $\frac{3}{2}.$

Hence,

$np=3\dots \dots \left(1\right)$

and

$\sqrt{npq}=\frac{3}{2}\dots \dots \left(2\right)$

By squaring equation $\left(2\right)$, we get

$npq=\frac{9}{4}\dots \dots \left(3\right)$

By dividing $\left(3\right)$ by $\left(1\right)$, we get

$$\begin{gathered} \frac{npq}{np}=\frac{{\displaystyle \frac{9}{4}}}{3} \\ \Rightarrow q=\frac{3}{4} \end{gathered}$$

Substituting the value of $q$ in $p+q=1$, we get

$$\begin{gathered} p=1-q \\ =1-\frac{3}{4} \\ =\frac{1}{4} \end{gathered}$$

Now, finding the value of $n$by substituting the value of $p$in the equation of mean.

$\begin{array}{rcl}np& =& 3\\ n& =& \frac{3}{p}\\ & =& \frac{3}{{\displaystyle \frac{1}{4}}}\\ \therefore n& =& 12\end{array}$

Step-2 : Probability distribution:

We know that the probability distribution of a binomial distribution with the parameters $n,p,q$ is ${\left(p+q\right)}^n$.

Here, $n=12,p=\frac{1}{4},q=\frac{3}{4}$.

Therefore, the required probability distribution is

$$\begin{gathered} {\left(q+p\right)}^n \\ ={\left(\frac{3}{4}+\frac{1}{4}\right)}^{12} \end{gathered}$$

Hence, option (A) is the correct answer.