The probability distribution of X is:
| x | 0 | 1 | 2 | 3 | 4 |
| P(X = x) | 0.1 | 2k | k | k | 2k |
Then var(X) =
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{159}{80}\)
The probabilities add up to 1:
\(0.1 + 2k + k + k + 2k = 1\)
\(6k = 0.9\)
\( \displaystyle k = 0.15 = \frac{3}{20} \)
The mean:
\(E(X) = 0(0.1) + 1(2k) + 2(k) + 3(k) + 4(2k)\)
\(= 15k\)
\( \displaystyle = 15 \times \frac{3}{20} = \frac{9}{4} \)
The mean of the squares:
\(E(X^2) = 0(0.1) + 1(2k) + 4(k) + 9(k) + 16(2k)\)
\(= 47k\)
\( \displaystyle = 47 \times \frac{3}{20} = \frac{141}{20} \)
The variance is \(E(X^2) - [E(X)]^2\):
\( \displaystyle \mathrm{var}(X) = \frac{141}{20} - \frac{81}{16} \)
\( \displaystyle = \frac{564 - 405}{80} = \frac{159}{80} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3}{20}\) — Incorrect. This is the value of k, not the variance.
Option (2) \(\displaystyle\frac{9}{4}\) — Incorrect. This is the mean E(X).
Option (3) \(\displaystyle\frac{141}{20}\) — Incorrect. This is \(E(X^2)\); the square of the mean has not been subtracted.