Probability density functions
Problem 5.706 · 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 \( \displaystyle P(0 \le X \le \frac{1}{2}) \) and the mean of \( \displaystyle X \).
- A density integrates to 1 over its whole range.
- \[ \int\limits_{0}^{1} x^{2}\, dx = \frac{1}{3} \]The integral without k.✓ Proved
- \[ 3 \cdot 1 \cdot \frac{1}{3} = 1 \]So k = 3: k times that integral is 1.✓ Proved
- \[ \int\limits_{0}^{\frac{1}{2}} 3 x^{2}\, dx = \frac{1}{8} \]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 \( k = 3,\ P = \frac{1}{8},\ \mu = \frac{3}{4} \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | 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: fail (error) — The probability over [0,1/2] is ∫₀^{1/2}3x²dx=3/8, not 1/8. The mean calculation is correct, but the probability step is wrong.qwen3.6:27b-mlx: pass — The solution correctly identifies the normalization condition, calculates k, computes the probability over the specified interval, and calculates the mean using the correct definition. All steps are logically sound and algebraically verified.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-20 — The solution correctly identifies the normalization condition, calculates k, computes the probability over the specified interval, and calculates the mean using the correct definition. All steps are logically sound and algebraically verified.gpt-oss:20b: fail (error) 2026-10-20 — The probability over [0,1/2] is ∫₀^{1/2}3x²dx=3/8, not 1/8. The mean calculation is correct, but the probability step is wrong.qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly identifies the normalization condition, calculates k, computes the probability over the specified interval, and calculates the mean using the correct definition. All steps are logically sound and algebraically verified.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly identifies the normalization constant, calculates the probability over the specified interval, and computes the mean using the correct integrals and bounds.gpt-oss:20b: fail (error) 2026-10-19 — The probability over [0,1/2] is ∫₀^{1/2}3x²dx = 3·(1/8)=3/8, not 1/8. The mean calculation is fine.
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 2026-10-20 with SymPy 1.14.0.