Let $X$ be random variable with distribution given below. If the mean of $X$ is 2.3 and variance of

Question

Let $X$ be random variable with distribution given below. If the mean of $X$ is 2.3 and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to

Professor PC Thomas Classes · Answer

(781) : $\bar{X}=\sum x_{i} p_{i}$\ $=\frac{-2}{5}-a+1+\frac{4}{5}+6 b=2.3$\ $\Rightarrow 6 b-a=2.3-1-0.4=0.9=\frac{9}{10}$\ Also, $\Sigma p\left(x_{i}ight)=1$\ $\Rightarrow \frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$\ $\Rightarrow a+b=\frac{2}{3}-\frac{2}{5}=\frac{4}{15}$\ Adding (i) and (ii), we have\ $7 b=\frac{9}{10}+\frac{4}{15} \Rightarrow b=\frac{1}{6}$\ $	herefore \quad a=\frac{4}{15}-\frac{1}{6}=\frac{1}{10}$\ Now, $\sigma^{2}=\sum p_{i} x_{i}^{2}-\bar{X}^{2}$\ $=\frac{4}{5}+\frac{1}{10} 	imes 1+\frac{1}{3} 	imes 9+\frac{1}{5} 	imes 16+\frac{1}{6} 	imes 36-(2.3)^{2}$\ $=\frac{4}{5}+\frac{1}{10}+3+\frac{16}{5}+6-(2.3)^{2}$\ $=13.1-5.29=7.81$\ $	herefore \quad 100 \sigma^{2}=781$\