If the mean of X is 2.3 and variance of X is ^2, then 100^2 is equal to Answer: 781

Question

If the mean of X is 2.3 and variance of X is $\sigma^2$, then $100\sigma^2$ is equal to

Answer: 781

Accepted Answer

Given distribution has probabilities involving $a, b$; mean condition:
$$-\frac25-a+1+\frac45+6b = 2.3 \implies 6b-a=0.9 \quad \text{...(i)}$$

Sum of probabilities $=1$: $\dfrac15+a+\dfrac13+\dfrac15+b=1 \implies a+b=\dfrac{4}{15}$ ...(ii)

Solving: $a=\dfrac{1}{10}, b=\dfrac16$

$$\sigma^2 = \sum P_ix_i^2-\mu^2 = \frac15(4)+\frac{1}{10}(1)+\frac13(9)+\frac15(16)+\frac16(36)-(2.3)^2 = 7.81$$

$$100\sigma^2 = 781$$

Solutions · 1