Q80 / 160 · AP EAMCET 2025 (26 MAY SHIFT 1) · Mathematics
The probability distribution of a random variable X is as follows. Then the mean of X is $\mathrm{X} = \mathrm{x}_{\mathrm{i}}$ $-2$ $-1$ $0$ $1$ $2$ $\mathrm{P}\left(\mathrm{X} = \mathrm{x}_{\mathrm{i}}\right)$ $\mathrm{k}^2 / 3$ $\mathrm{k}^2$ $2 \mathrm{k}^2 / 3$ $k / 2$ $k / 2$
| $\mathrm{X} = \mathrm{x}_{\mathrm{i}}$ | $-2$ | $-1$ | $0$ | $1$ | $2$ |
|---|---|---|---|---|---|
| $\mathrm{P}\left(\mathrm{X} = \mathrm{x}_{\mathrm{i}}\right)$ | $\mathrm{k}^2 / 3$ | $\mathrm{k}^2$ | $2 \mathrm{k}^2 / 3$ | $k / 2$ | $k / 2$ |
- Correct answerA$\frac{1}{3}$
- B$\frac{1}{5}$
- C$\frac{11}{2}$
- D$\frac{13}{2}$
Hence, the correct answer is option: 'A'.