Questions: Continuous Uniform Distribution

Question 1

Let X ~ Uniform[0, 4]. A student computes P(X = 2) = f(2) = 1/4, using the PDF value directly. What error has the student made?

• A. The PDF value should be evaluated at the midpoint (0+4)/2 = 2, not at 2 directly
• B. The student confused PDF density with probability — P(X = 2) = 0 for any specific point in a continuous distribution
• C. The correct answer is 1/2 because x = 2 is the mean of the distribution
• D. The calculation is correct; the PDF value at a point equals the probability of that point

Answer: (B)

For any continuous distribution, P(X = c) = 0 for every specific value c — the probability of hitting an exact point is zero. The PDF f(x) = 1/(b−a) gives probability *density*, not probability. Probability requires integrating the PDF over an interval (an area, not a point). The common confusion is treating f(x) as if it were a probability mass function from a discrete distribution.

Question 2

X ~ Uniform[2, 8]. What is P(3 ≤ X ≤ 5)?

• A. 1/6 — the PDF value is 1/6, so that is the probability for any subinterval
• B. 1/3 — the length of [3, 5] divided by the total length of [2, 8] gives 2/6
• C. 1/4 — because the subinterval [3, 5] starts one unit in from the left boundary
• D. 2/8 — the length of the subinterval divided by the right endpoint

Answer: (B)

P(3 ≤ X ≤ 5) = (5 − 3)/(8 − 2) = 2/6 = 1/3. The probability equals the length of the target interval divided by the total interval length — a rectangle with height 1/6 and width 2. The PDF value (1/6) is not itself the probability; it must be multiplied by the interval length. This is the continuous analogue of classical probability as a ratio.

Question 3

For X ~ Uniform[a, b], the mean is (a + b)/2 because the distribution is symmetric about its midpoint.

• T. True
• F. False

Answer: True

The uniform distribution assigns equal density to every point in [a, b], making the distribution perfectly symmetric around (a + b)/2. By symmetry, the expected value must be the midpoint. This can also be confirmed by evaluating ∫ x · (1/(b−a)) dx from a to b, which yields (a + b)/2.

Question 4

The PDF value f(x) = 1/(b − a) represents the probability that X takes the value x in a continuous uniform distribution.

• T. True
• F. False

Answer: False

This is the central misconception about continuous distributions. The PDF gives density — the rate of probability accumulation per unit length — not probability at a point. The probability that X equals any single value is always zero for a continuous distribution. Probability is the area under the PDF curve over an interval, not the height at a point.

Question 5

Why is the continuous uniform distribution described as the 'raw material' for generating other continuous random variables? What property makes this possible?

Model answer: The probability integral transform states that if X has any continuous CDF F, then F(X) ~ Uniform(0, 1). Inverting this, if U ~ Uniform(0, 1), then F⁻¹(U) has CDF F — so any continuous distribution can be generated by applying an inverse CDF to a uniform random variable. Computer random number generators produce Uniform(0,1) samples, then transform them into whatever distribution is needed.

This makes the uniform distribution foundational to simulation and Monte Carlo methods. Every sampling algorithm ultimately relies on generating uniform randomness and transforming it. The uniform distribution is 'maximally uninformative' over an interval, and that property is what allows any structure to be imposed on it through transformation.