If the probability distribution of X is :
| X | 2 | 3 | 4 | 5 | 6 |
| P(X) | $\frac{1}{15}$ | $\frac{2}{15}$ | $\frac{3}{15}$ | $\frac{4}{15}$ | $\frac{5}{15}$ |
Then variance is equal to :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{14}{9}$
The expected Mean Value is,
$E(X)=∑X.P(X)$
$=2×\frac{1}{15}+3×\frac{2}{15}+4×\frac{3}{15}+5×\frac{4}{15}+6×\frac{5}{15}$
$=\frac{70}{15}$
$E(X^2)=∑X^2.P(X)$
$=2^2×\frac{1}{15}+3^2×\frac{2}{15}+4^2×\frac{3}{15}+5^2×\frac{4}{15}+6^2×\frac{5}{15}$
$=\frac{350}{15}$
Variance, $σ^2=E(X^2)-[E(X)]^2$
$=\frac{350}{15}-\frac{4900}{225}$
$=\frac{5250-4900}{225}=\frac{350}{225}=\frac{14}{9}$