Probability density functions
Problem 5.666 · medium
Find \( \displaystyle k \) so that \( \displaystyle f(x) = k\left(x^{2}\right) \) for \( \displaystyle 0 \le x \le 1 \) (and 0 elsewhere) is a probability density. Then find \( \displaystyle P(0 \le X \le 1) \) and the mean of \( \displaystyle X \).
- A density integrates to 1 over its whole range.Reviewed
- \[ \int\limits_{0}^{1} x^{2}\, dx = \frac{1}{3} \]The integral without k.✓ Proved
- \[ 3 \cdot 1 \cdot \frac{1}{3} = 1 \]So k = 3: k times that integral is 1.✓ Proved
- \[ \int\limits_{0}^{1} 3 x^{2}\, dx = 1 \]The probability is the area over the interval.✓ Proved
- \[ \int\limits_{0}^{1} 3 x^{3}\, dx = \frac{3}{4} \]The mean, ∫ x f(x) dx.✓ Proved
Answer \( k = 3,\ P = 1,\ \mu = \frac{3}{4} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | mpmath quadrature for the normalizing constant, the probability and the mean |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the normalization constant, calculates the probability over the support (which is 1 by definition), and computes the mean. The logic is sound and the steps are clearly justified.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the normalization constant, calculates the probability over the support (which is 1 by definition), and computes the mean. The logic is sound and the steps are clearly justified.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the normalization constant, calculates the probability over the support (which is 1 by definition), and computes the mean using the correct formula. The steps are logically sound and algebraically verified.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the normalization constant, calculates the total probability, and computes the mean using standard definitions.gpt-oss:20b: pass 2026-10-15
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/probability_density, checked 2026-10-18 with SymPy 1.14.0.