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Paper 4 (9164, Section A) · November 2004 · Continuous Random Variables

A number X is randomly selected from the interval (−π, π)(-\pi,\ \pi), so that X is uniformly distributed over that interval. Which of these is the cumulative distribution function of X?

AF(x)=0F(x)=0 for x≤−πx\le-\pi; F(x)=x2πF(x)=\dfrac{x}{2\pi} for −π<x<π-\pi<x<\pi; F(x)=1F(x)=1 for x≥πx\ge\pi
BF(x)=0F(x)=0 for x≤−πx\le-\pi; F(x)=x+2π2πF(x)=\dfrac{x+2\pi}{2\pi} for −π<x<π-\pi<x<\pi; F(x)=1F(x)=1 for x≥πx\ge\pi
CF(x)=0F(x)=0 for x≤−πx\le-\pi; F(x)=x+π2πF(x)=\dfrac{x+\pi}{2\pi} for −π<x<π-\pi<x<\pi; F(x)=1F(x)=1 for x≥πx\ge\pi
DF(x)=0F(x)=0 for x≤−πx\le-\pi; F(x)=12πF(x)=\dfrac{1}{2\pi} for −π<x<π-\pi<x<\pi; F(x)=1F(x)=1 for x≥πx\ge\pi

Explanation

The interval has length 2π2\pi, so the density is f(x)=12πf(x)=\dfrac{1}{2\pi} on it. Integrating from the lower end, F(x)=∫−πx12π dt=x+π2πF(x)=\displaystyle\int_{-\pi}^{x}\frac{1}{2\pi}\,dt=\frac{x+\pi}{2\pi} for −π<x<π-\pi<x<\pi, which rises from 0 at x=−πx=-\pi to 1 at x=πx=\pi. The middle piece 12π\frac{1}{2\pi} is the density itself, not the distribution function; x2π\frac{x}{2\pi} gives −12-\frac12 rather than 0 at x=−πx=-\pi; and x+2π2π\frac{x+2\pi}{2\pi} gives 12\frac12 at x=−πx=-\pi and 32\frac32 at x=πx=\pi, so neither of those runs from 0 to 1.

Derived from ZIMSEC Statistics Paper 4, November 2004, Q1

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