Probability density functions
Problem 5.606 · medium
Find \( \displaystyle k \) so that \( \displaystyle f(x) = k\left(e^{- \frac{x}{3}}\right) \) for \( \displaystyle x \ge 0 \) (and 0 elsewhere) is a probability density. Then find \( \displaystyle P(0 \le X \le 3) \) and the mean of \( \displaystyle X \).
- A density integrates to 1 over its whole range.Reviewed
- \[ \int\limits_{0}^{\infty} e^{- \frac{x}{3}}\, dx = 3 \]The integral without k.✓ Proved
- \[ 1 \cdot \frac{1}{3} \cdot 3 = 1 \]So k = 1/3: k times that integral is 1.✓ Proved
- \[ \int\limits_{0}^{3} \frac{e^{- \frac{x}{3}}}{3}\, dx = 1 - e^{-1} \]The probability is the area over the interval.✓ Proved
- \[ \int\limits_{0}^{\infty} \frac{x e^{- \frac{x}{3}}}{3}\, dx = 3 \]The mean, ∫ x f(x) dx.✓ Proved
Answer \( k = \frac{1}{3},\ P = 1 - e^{-1},\ \mu = 3 \)
✓ 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, computes the constant k, calculates the probability over the specified interval, and determines the mean using the standard definitions for a continuous random variable.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly identifies the normalization condition, computes the constant k, calculates the probability over the specified interval, and determines the mean using the standard definitions for a continuous random variable.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-14gpt-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. All steps are logically sound and mathematically correct.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.