Consider a probability distribution given by the density function \(P(x)\).
\[P(x) = \begin{cases} Cx^2, & \text{for } 1 \le x \le 4 \\ 0, & \text{for } x \lt 1 \text{ or } x \gt 4 \end{cases}\]
The probability that \(x\) lies between 2 and 3, i.e., \(P(2 \le x \le 3)\) is __________. (rounded off to three decimal places)
Answer
0.3
Explanation
The total probability must be 1, which fixes \(C\):
\[\int_1^4 Cx^2\,dx = C \cdot \frac{4^3 - 1^3}{3} = 21C = 1 \;\Rightarrow\; C = \frac{1}{21}\]
\[P(2 \le x \le 3) = \frac{1}{21}\int_2^3 x^2\,dx = \frac{1}{21} \cdot \frac{3^3 - 2^3}{3} = \frac{19}{63} \approx 0.3016\]
Rounded to three decimal places, this is 0.302.
Answer: 0.302