AP EAMCET · Maths · Probability
The probability distribution of a random variable X is as follows. Then the mean of X is
| \(\mathrm{X} = \mathrm{x}_{\mathrm{i}}\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) |
|---|---|---|---|---|---|
| \(\mathrm{P}\left(\mathrm{X} = \mathrm{x}_{\mathrm{i}}\right)\) | \(\mathrm{k}^2 / 3\) | \(\mathrm{k}^2\) | \(2 \mathrm{k}^2 / 3\) | \(k / 2\) | \(k / 2\) |
- A \(\frac{1}{3}\)
- B \(\frac{1}{5}\)
- C \(\frac{11}{2}\)
- D \(\frac{13}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\( \sum P(X=x_i) = 1 \) \( \frac{k^2}{3} + k^2 + \frac{2k^2}{3} + \frac{k}{2} + \frac{k}{2} = 1 \) \( 2k^2 + k - 1 = 0 \) \( (2k-1)(k+1) = 0 \) \( k = \frac{1}{2} \) (since \( k \ne -1 \) for valid probabilities) \( E[X] = \sum x_i P(X=x_i) \)…
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