Probability density functions
Problem 5.593 · medium
Find \( \displaystyle k \) so that \( \displaystyle f(x) = k\left(x\right) \) for \( \displaystyle 0 \le x \le 4 \) (and 0 elsewhere) is a probability density. Then find \( \displaystyle P(1 \le X \le \frac{5}{2}) \) and the mean of \( \displaystyle X \).
- A density integrates to 1 over its whole range.
- \[ \int\limits_{0}^{4} x\, dx = 8 \]The integral without k.✓ Proved
- \[ 1 \cdot \frac{1}{8} \cdot 8 = 1 \]So k = 1/8: k times that integral is 1.✓ Proved
- \[ \int\limits_{1}^{\frac{5}{2}} \frac{x}{8}\, dx = \frac{21}{64} \]The probability is the area over the interval.✓ Proved
- \[ \int\limits_{0}^{4} \frac{x^{2}}{8}\, dx = \frac{8}{3} \]The mean, ∫ x f(x) dx.✓ Proved
Answer \( k = \frac{1}{8},\ P = \frac{21}{64},\ \mu = \frac{8}{3} \)
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: passqwen3.6:27b-mlx: fail (error) — The solution calculates the second moment (E[X^2]) in step 5 but labels it as the mean. The mean requires integrating x*f(x), not x^2*f(x).
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The solution calculates the second moment (E[X^2]) in step 5 but labels it as the mean. The mean requires integrating x*f(x), not x^2*f(x).gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The solution calculates the second moment (E[X^2]) instead of the mean (E[X]). The mean is defined as the integral of x*f(x), but the equation integrates x^2*f(x).gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the normalization condition, calculates the constant k, computes the probability over the specified interval, and determines the mean using the correct definition and bounds.gpt-oss:20b: pass 2026-10-14
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 2026-10-17 with SymPy 1.14.0.