AP EAMCET · Maths · Probability
The probability distribution of a discrete random variable X is given below
| X = x | -1 | 0 | 1 | 2 |
|---|---|---|---|---|
| P(X = x) | \(\frac{1}{3}\) | \(\frac{1}{6}\) | \(\frac{1}{6}\) | \(\frac{1}{3}\) |
Then the value of \(6 \Sigma\left(\mathrm{x}^2\right) \mathrm{P}(\mathrm{X}=\mathrm{x})-\operatorname{var}(\mathrm{X})=\)
- A \(\frac{113}{12}\)
- B \(\frac{151}{12}\)
- C \(\frac{19}{12}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{113}{12}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{E}[X^2] = (-1)^2\left(\frac{1}{3}\right) + (0)^2\left(\frac{1}{6}\right) + (1)^2\left(\frac{1}{6}\right) + (2)^2\left(\frac{1}{3}\right) = \frac{1}{3} + 0 + \frac{1}{6} + \frac{4}{3} = \frac{2+1+8}{6} = \frac{11}{6} \)…
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