Question

A binomial variate X has mean = 6 and variance = 2 the probability that 5 $\le$ X $\le$ 7 is

• A. $\frac{4}{2}$
• B. $\frac{1622}{6661}$
• C. $\frac{4672}{6561}$
• D. none

Answer 1

Answer: (C)

Here mean = np = 6 ; variance = npq =2
$\therefore q=\frac{2}{6}=\frac{1}{3};\therefore p=1-q=\frac{2}{3};\therefore n=9$
Now, probability of $5\le X\le 7$ is equal to $P\left(5\right)+P\left(6\right)+P\left(7\right)$
${=}^9{C}_5{p}^5{q}^4{+}^9{C}_6{p}^6{q}^3{+}^9{C}_7{p}^7{q}^2$