Consider a probability distribution given by the density function \(𝑃(𝑥)\).…
2025
Consider a probability distribution given by the density function \(𝑃(𝑥)\).
\(P(x) =
\begin{cases}
Cx^2, & \text{for } 1 \leq x \leq 4 \\
0, & \text{for } x < 1 \text{ or } x > 4
\end{cases}
\)
The probability that \(x\) lies between 2 and 3, i.e., \(𝑃(2 ≤ 𝑥 ≤ 3)\) is __________. (rounded off to three decimal places)
Answer: 0.302 — Normalize the density to find the constant C: Integrate the density over its support and set the total probability to 1: Compute ∫ from 1 to 4 of C x² dx = 1.…
Attempted by 21 students.
Show answer & explanation
Correct answer: 0.302
Normalize the density to find the constant C:
Integrate the density over its support and set the total probability to 1:
Compute ∫ from 1 to 4 of C x² dx = 1. This gives C · [x³/3]₁⁴ = 1.
Evaluate the bracket: x³/3 from 1 to 4 = (64/3) − (1/3) = 63/3 = 21. So C · 21 = 1, hence C = 1/21.
Now compute the probability that x lies between 2 and 3:
P(2 ≤ x ≤ 3) = ∫₂³ (1/21) x² dx = (1/21) · [x³/3]₂³.
Evaluate: [x³/3]₂³ = (27/3) − (8/3) = 19/3. So the probability is (1/21) · (19/3) = 19/63.
Numeric value rounded to three decimal places:
19/63 ≈ 0.302
A video solution is available for this question — log in and enroll to watch it.