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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability4 Marks
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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