Probability density functions
Problem 5.777 · medium
Find \( \displaystyle k \) so that \( \displaystyle f(x) = k\left(x^{2}\right) \) for \( \displaystyle 0 \le x \le 3 \) (and 0 elsewhere) is a probability density. Then find \( \displaystyle P(\frac{5}{2} \le X \le 3) \) and the mean of \( \displaystyle X \).
- A density integrates to 1 over its whole range.Reviewed
- \[ \int\limits_{0}^{3} x^{2}\, dx = 9 \]The integral without k.✓ Proved
- \[ 1 \cdot \frac{1}{9} \cdot 9 = 1 \]So k = 1/9: k times that integral is 1.✓ Proved
- \[ \int\limits_{\frac{5}{2}}^{3} \frac{x^{2}}{9}\, dx = \frac{91}{216} \]The probability is the area over the interval.✓ Proved
- \[ \int\limits_{0}^{3} \frac{x^{3}}{9}\, dx = \frac{9}{4} \]The mean, ∫ x f(x) dx.✓ Proved
Answer \( k = \frac{1}{9},\ P = \frac{91}{216},\ \mu = \frac{9}{4} \)
✓ 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 condition, calculates the constant 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 2026-11-06 — The solution correctly identifies the normalization condition, calculates the constant 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 2026-11-06qwen3.6:27b-mlx: pass 2026-11-05 — 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 integrals and bounds.gpt-oss:20b: pass 2026-11-05qwen3.6:27b-mlx: pass 2026-11-05 — The solution correctly applies the definition of a probability density function, calculates the normalization constant, and computes the probability and mean using the correct integrals and bounds.gpt-oss:20b: pass 2026-11-05
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 2026-11-06 with SymPy 1.14.0.