Statistics
Probability Distribution — Variance
nta_pyq_2024_apr
Grade 11
Question:
If the mean of the following probability distribution of a random variable $X$:
| $X$ | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| $P(X)$ | $a$ | $2a$ | $a+b$ | $2b$ | $3b$ |
is $\dfrac{46}{9}$, then the variance of the distribution is
$\dfrac{173}{27}$
$\dfrac{566}{81}$
$\dfrac{151}{27}$
$\dfrac{581}{81}$
Step-by-Step Solution
Key Concept: $\sum P_i=1\Rightarrow4a+6b=1$ ...(I). $E(X)=\frac{46}{9}\Rightarrow4a+20b=\frac{23}{9}$ ...(II). Subtracting: $14b=\frac{23}{9}-1=\frac{14}{9}\Rightarrow b=\frac{1}{9}$, $a=\frac{1}{12}$.
$a=\frac{1}{12}$, $b=\frac{1}{9}$. Variance $=\frac{566}{81}$.
Correct Answer: 2