The height of the probability density function f(x) of a random variable X that is uniformly distributed between a and b is:

Question

BlackTom AI · Accepted Answer

The question concerns the height of the probability density function (pdf) of a uniformly distributed random variable X over the interval [a, b].

For a continuous uniform distribution on [a, b], the pdf f(x) is constant across the interval and zero outside it. The constant value must be chosen so that the total probability over the interval equals 1:

∫ from a to b f(x) dx = 1. Since f(x) is constant on [a, b], let f(x) = c for x in [a, b]. Then c·(b − a) = 1, which yields c = 1/(b − a).

Therefore, the height of the pdf on the interval [a, b] is 1/(b − a). Outside of [a, b], the pdf is zero, but the question specifically asks for the height on the interval where the density is nonzero.

This reasoning aligns with the normalization condition of a pdf and the defining property of a uniform distribution. Any alternative value would fail to integrate to 1 over [a, b] or would violate the requirement that the density is constant on the interval.

Note: The exact numeric value depends on the labeled parameters a and b with a < b; if a and b are given specific numbers, you would substitute them into 1/(b − a).

COMM111_S2 Quiz 4: Continuous Distributions

The height of the probability density function f(x) of a random variable X that is uniformly distributed between a and b is:

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