Find the binomial distribution whose mean is 5 and variance 10 3 … - Vidyadip

Question

Find the binomial distribution whose mean is 5 and variance $\frac{10}{3}.$

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Answer

Let n and p be the parameters of binomial distribution.
Given, $\text{Mean = np}=5\dots(1)$
$\text{Variance = npq}=\frac{10}{3}\dots(2)$
Dividing (2) by (1)
$\frac{\text{npq}}{\text{np}}=\frac{\frac{10}{3}}{5}$
$\text{q}=\frac{2}{3}$
So, $\text{p}=1-\text{q}$ [Since p + q = 1]
$=1-\frac{2}{3}$
$\text{p}=\frac{1}{3}$
Put the value of p in equation (1),
$\text{np}=5$
$\text{n}=5\times3$
$\text{n}=15$
Hence, the binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^{15}\text{c}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{15-\text{r}}$
$\text{r}=0,1,2,\dots15$

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