Question
If a random variable X follows a binomial distribution with mean 3 and variance 3/2, find P (X ≤ 5).
✓
Answer
$\text{Mean(np)}=3$ and $\text{Variance(npq)}=\frac{3}2{}$
$\therefore\text{q}=\frac{1}{2}$
and $\text{p}=1-\frac{1}{2}$
$\text{n}=\frac{\text{Mean}}{\text{p}}$
$\Rightarrow\text{n}=6$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^{6}\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}},\text{r}=0,1,2\dots6$
$=\frac{\text{ }^6\text{C}_{\text{r}}}{2^6}$
$\therefore\text{P(X}\leq5)=1-\text{P(X}=6)$
$=1-\frac{1}{64}$
$=\frac{63}{64}$
$\therefore\text{q}=\frac{1}{2}$
and $\text{p}=1-\frac{1}{2}$
$\text{n}=\frac{\text{Mean}}{\text{p}}$
$\Rightarrow\text{n}=6$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^{6}\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{6-\text{r}},\text{r}=0,1,2\dots6$
$=\frac{\text{ }^6\text{C}_{\text{r}}}{2^6}$
$\therefore\text{P(X}\leq5)=1-\text{P(X}=6)$
$=1-\frac{1}{64}$
$=\frac{63}{64}$
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