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Probability density functions

Problem 5.600 · medium

Find \( \displaystyle k \) so that \( \displaystyle f(x) = k\left(x\right) \) for \( \displaystyle 0 \le x \le 2 \) (and 0 elsewhere) is a probability density. Then find \( \displaystyle P(0 \le X \le 2) \) and the mean of \( \displaystyle X \).

A density integrates to 1 over its whole range.
Reviewed
\[ \int\limits_{0}^{2} x\, dx = 2 \]
The integral without k.✓ Proved
\[ 1 \cdot \frac{1}{2} \cdot 2 = 1 \]
So k = 1/2: k times that integral is 1.✓ Proved
\[ \int\limits_{0}^{2} \frac{x}{2}\, dx = 1 \]
The probability is the area over the interval.✓ Proved
\[ \int\limits_{0}^{2} \frac{x^{2}}{2}\, dx = \frac{4}{3} \]
The mean, ∫ x f(x) dx.✓ Proved

Answer \( k = \frac{1}{2},\ P = 1,\ \mu = \frac{4}{3} \)

✓ 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: pass
qwen3.6:27b-mlx: pass — The solution correctly identifies the normalization condition, calculates the constant k, and computes the probability and mean using the correct integrals and bounds.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly identifies the normalization condition, calculates the constant k, and computes the probability and mean using the correct integrals and bounds.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the normalization constant, calculates the probability over the support (which is 1 by definition for a valid PDF), and computes the mean using the correct formula.
gpt-oss:20b: pass 2026-10-14
qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the normalization condition, calculates k, and computes the probability and mean using the correct formulas 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.