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In a binomial distribution, the mean is 4 and the variance is 3 . What is the mode?
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The correct answer is:
4
As give, $n p=4$ and $n p q=3$ [where $\mathrm{p}$ is the probability of success and $\mathrm{q}$ is the probability of failure for an event to occur, and 'n' is the number of trials
$\Rightarrow \mathrm{q}=\frac{\mathrm{npq}}{\mathrm{np}}=\frac{3}{4}$
Also, $\mathrm{p}=1-\mathrm{q}=1-\frac{3}{4}=\frac{1}{4}$
$\mathrm{n}=16$
In a binomial distribution, the value of $\mathrm{r}$ for which $\mathrm{P}(\mathrm{X}=\mathrm{r})$ is maximum is the mode of binomial distribution.
hence, $(\mathrm{n}+1) \mathrm{p}-1 \leq \mathrm{r} \leq(\mathrm{n}+1) \mathrm{p}$
$\Rightarrow \frac{17}{4}-1 \leq \mathrm{r} \leq \frac{17}{4}$
$\Rightarrow \frac{13}{4} \leq \mathrm{r} \leq \frac{17}{4}$
$\Rightarrow 3.25 \leq \mathrm{r} \leq 4.25$
$\Rightarrow \mathrm{r}=4$
$\Rightarrow \mathrm{q}=\frac{\mathrm{npq}}{\mathrm{np}}=\frac{3}{4}$
Also, $\mathrm{p}=1-\mathrm{q}=1-\frac{3}{4}=\frac{1}{4}$
$\mathrm{n}=16$
In a binomial distribution, the value of $\mathrm{r}$ for which $\mathrm{P}(\mathrm{X}=\mathrm{r})$ is maximum is the mode of binomial distribution.
hence, $(\mathrm{n}+1) \mathrm{p}-1 \leq \mathrm{r} \leq(\mathrm{n}+1) \mathrm{p}$
$\Rightarrow \frac{17}{4}-1 \leq \mathrm{r} \leq \frac{17}{4}$
$\Rightarrow \frac{13}{4} \leq \mathrm{r} \leq \frac{17}{4}$
$\Rightarrow 3.25 \leq \mathrm{r} \leq 4.25$
$\Rightarrow \mathrm{r}=4$
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