Question
Determine the binomial distribution whose mean is 20 and variance 16.
✓
Answer
Given that,
$\text{Mean = np}=20\dots(1)$
$\text{Variance = npq}=16\dots(2)$
Let n and p be the parameters of distribution dividing equation (2) by (1)
$\frac{\text{npq}}{\text{np}}=\frac{16}{20}$
$\text{q}=\frac{4}{5}$
So, $\text{p}=1-\text{q}$ [Since p + q = 1]
$=1-\frac{4}{5}$
$\text{p}=\frac{1}{5}$
Put p in equation (1),
$\text{np}=20$
$\text{n}\big(\frac{1}{5}\big)=20$
$\text{n}=20\times5$
$\text{n}=100$
So, binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^{100}\text{c}_{\text{r}}\big(\frac{1}{5}\big)^{\text{r}}\big(\frac{4}{5}\big)^{100-\text{r}}$
$\text{r}=0,1,2,3,\dots100$
$\text{Mean = np}=20\dots(1)$
$\text{Variance = npq}=16\dots(2)$
Let n and p be the parameters of distribution dividing equation (2) by (1)
$\frac{\text{npq}}{\text{np}}=\frac{16}{20}$
$\text{q}=\frac{4}{5}$
So, $\text{p}=1-\text{q}$ [Since p + q = 1]
$=1-\frac{4}{5}$
$\text{p}=\frac{1}{5}$
Put p in equation (1),
$\text{np}=20$
$\text{n}\big(\frac{1}{5}\big)=20$
$\text{n}=20\times5$
$\text{n}=100$
So, binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X = r})=\text{ }^{100}\text{c}_{\text{r}}\big(\frac{1}{5}\big)^{\text{r}}\big(\frac{4}{5}\big)^{100-\text{r}}$
$\text{r}=0,1,2,3,\dots100$
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