A random variable X has the following probability distribution:
| X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(X) | 0 | \(k\) | \(2k\) | \(2k\) | \(3k\) | \(k^2\) | \(2k^2\) | \(7k^2 + k\) |
The value of \(P(4 \lt x \lt 7)\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{100}\)
The probabilities add up to 1:
\(0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1\)
\(10k^2 + 9k - 1 = 0\)
\((10k - 1)(k + 1) = 0\)
A probability cannot be negative, so \(k = -1\) is rejected:
\( \displaystyle k = \frac{1}{10} \)
The inequalities are strict, so only X = 5 and X = 6 lie in the range:
\(P(4 \lt X \lt 7) = P(5) + P(6) = k^2 + 2k^2 = 3k^2\)
\( \displaystyle 3k^2 = 3 \times \frac{1}{100} = \frac{3}{100} \)