Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability4 Marks
Question
Find the binomial distribution whose mean is 5 and variance $\frac{10}{3}.$
✓
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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