For the given probability distribution,
| X | −2 | −1 | 0 | 1 | 2 |
| P(X) | 1/K | 2/K | 3/K | 1/K | 2/K |
which of the following statements are correct?
(A) Mean of X is 0
(B) Mean of X is 1/9
(C) Variance of X is 5/3
(D) Variance of X is 134/81
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (D) only
The probabilities add up to 1:
\( \displaystyle \frac{1+2+3+1+2}{K}=1\ \Rightarrow\ K=9 \)
\( \displaystyle E(X)=\sum xP(x) \)
\( \displaystyle =\frac{(-2)(1)+(-1)(2)+0(3)+1(1)+2(2)}{9} \)
\( \displaystyle =\frac{-2-2+0+1+4}{9}=\frac{1}{9} \)
\( \displaystyle E(X^2)=\sum x^2P(x) \)
\( \displaystyle =\frac{4(1)+1(2)+0(3)+1(1)+4(2)}{9} \)
\( \displaystyle =\frac{15}{9}=\frac{5}{3} \)
\( \displaystyle \mathrm{Var}(X)=E(X^2)-\left[E(X)\right]^2 \)
\( \displaystyle =\frac{5}{3}-\frac{1}{81} \)
\( \displaystyle =\frac{135-1}{81}=\frac{134}{81} \)
So the mean is \( \frac{1}{9} \) and the variance is \( \frac{134}{81} \).
Checking the statements
(A) Mean of X is 0 — Incorrect. The mean is \( \frac{1}{9} \), not 0.
(B) Mean of X is 1/9 — Correct. The mean is \( \frac{1}{9} \).
(C) Variance of X is 5/3 — Incorrect. \( \frac{5}{3} \) is \( E(X^2) \); the variance is \( \frac{134}{81} \).
(D) Variance of X is 134/81 — Correct. The variance is \( \frac{134}{81} \).