Determine the binomial distribution whose mean is 9 and variance … - Vidyadip

Question

Determine the binomial distribution whose mean is 9 and variance $\frac{9}{4}.$

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Answer

Let X denote the variance with parameters n and p $\text{p + q}=1$ $\text{q}=1-\text{p}$Given,
$\text{Mean = np} =9\dots(1)$ $\text{variance = npq}=\frac{9}{4}\dots(2)$ $\frac{\text{npq}}{\text{np}}=\frac{\frac{9}{4}}{9}$ [By diving (1) by (2)] $\text{q}=\frac{1}{4}$ So, $\text{p}=1-\text{q}$ $=1-\frac{1}{4}$ $\text{p}=\frac{3}{4}$ Put p in equation (1), $\text{n}\big(\frac{3}{4}\big)=9$ $\Rightarrow\text{n}=\frac{36}{3}$ So, $\text{n}=12$ The distribution is given by $=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}(\text{q})^{\text{n}-\text{r}}$ $\text{P(X = r})=\text{ }^{12}\text{c}_{\text{r}}\big(\frac{3}{4}\big)^{\text{r}}\big(\frac{1}{4}\big)^{12-\text{r}}$ $\text{for r}=0,1,2,\dots12$

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