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
| x | 0 | 1 | 2 | 3 | 4 |
| P(X = x) |
0.1 |
2k | k | k | 2k |
∑P(Xi) = 1
⇒ 0.1 + 2k + k + k + 2k = 1
⇒ 6k = 0.7
so $k = \frac{0.9}{6} = \frac{9}{60} = \frac{3}{20}$
so var(X) = E(X2) - E2(X)
= 02 × 0.1 + 12 × 2k + 22 × k + 32 × k + 42 × 2k - (0 × 0.1 + 1 × 2k + 2 × k + 3 × k + 4 × 2k)2
= 2k + 4k + 9k + 32k - (2k + 2k + 3k + 8k)2
= 47 - (15k)2
= 47k - 225k2
= $47 × \frac{3}{20} - 225 × \frac{9}{400}$
= $\frac{141}{20} - \frac{81}{16}$
= $\frac{564 - 405}{80}$
= $\frac{159}{80}$