Binomial Distribution — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSBinomial Distribution3 Marks
Question
The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.
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Answer
Given that mean, i.e. $\text{np}=20\dots(1)$
and standard deviation, i.e. $\text{npq}=4$
$\sqrt{\text{npq}}=4$
$\Rightarrow\text{npq}=16\dots(2)$
Dividing eq. (2) by eq. (1), we get
$\text{q}=\frac{16}{20}=\frac{4}{5}$
and $\text{p}=\frac{1}{5};$
$\therefore\text{n}=\frac{\text{Mean}}{\text{p}}=100$
$\text{P(X = r})=\text{ }^{100}\text{C}_{\text{r}}\big(\frac{1}5{})^{\text{r}}\big(\frac{4}{5}\big)^{100-\text{r}},\text{r}=0,1,2\dots100$
Therefore, the parameters are $\text{n}=100$ and $\text{p}=\frac{1}{5}$
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