Q. A random variable $X$ can attain only the value 1, 2, 3, 4, 5 with respective probabilities $k,2k,3k,2k,k$. If $m$ is the mean of the probability distribution, then$(k,m)$is equal to
A) $\left( 3,\frac{1}{9} \right)$
B) $\left( \frac{1}{9},3 \right)$
C) $\left( \frac{1}{8},4 \right)$
D) $(1,\ 3)$
Explanation
(b)
We know sum of probability distribution is 1
$\therefore\ k + 2k + 3k + 2k + k = 1$
$\Rightarrow \ k = \frac{1}{9}$
$\therefore Mean,m = \sum_{i = 1}^{5}{P_{i}x_{i}}$
$= k(1) + 2k(2) + 3k(3) + 2k(4) + k(5)$
$= k(1 + 4 + 9 + 8 + 5) = \frac{1}{9} \times 27 = 3$
$\therefore\ (k,m) = \left( \frac{1}{9},3 \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.