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Modelling with the continuous uniform distribution

Edexcel IAL Mathematics · WST02 Statistics 2

The uniform distribution models situations where every value in a range is equally likely: rounding errors, where a rod snaps, or how long you wait for something that runs to a fixed timetable.

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Typical models

Rounding errors: a length measured to the nearest cm has error E ~ U[-0.5, 0.5].
A random cut: a 20 cm rod cut at a random point X from one end: X ~ U[0, 20].
Waiting time: buses every 15 minutes and you arrive at a random time: wait W ~ U[0, 15].

Examiners report students choosing the wrong interval (for example U[0, 42] instead of U[21, 42]); read the question carefully to find the actual range of possible values.

Shorter and longer pieces

For a 20 cm rod cut at X ~ U[0, 20], the shorter piece is less than 4 cm if X < 4 or X > 16: probability 8/20 = 2/5. 'The longer piece is more than 3 times the shorter' means the shorter piece is less than 5 cm: X < 5 or X > 15, probability 1/2. Sketch the rod.

Functions and combining models

For the area of a square with side X ~ U[1, 4], E(X²) = Var(X) + [E(X)]² = 9/12 + 2.5² = 7, not 2.5².

You may combine with the binomial: if P(|E| > 0.3) = 0.4 for one rounding error, the number of 5 independent errors with |E| > 0.3 is B(5, 0.4).

Worked example

A 20 cm rod is cut at a random point. Find the probability that the shorter piece is less than 4 cm.

X ~ U[0, 20] is the distance of the cut from one end.
Shorter piece < 4 when X < 4 or X > 16.
P = 4/20 + 4/20

Answer: 2/5

Key idea: Find the real range first, sketch, then use rectangle areas. For E(X²) use Var(X) + [E(X)]².

Check yourself

A 20 cm rod is cut at a random point. Find the probability that the shorter piece is less than 4 cm. Give your answer as a fraction in its simplest form.
Show answer
2/5. X < 4 or X > 16: 8/20 = 2/5.
A 20 cm rod is cut at a random point. Find the probability that the longer piece is more than three times the length of the shorter piece. Give your answer as a fraction in its simplest form.
Show answer
1/2. Longer > 3 × shorter means shorter < 5: X < 5 or X > 15, so 10/20 = 1/2.
Buses leave every 15 minutes and a passenger arrives at a random time. Find the probability that she waits more than 10 minutes. Give your answer as a fraction in its simplest form.
Show answer
1/3. W ~ U[0, 15]; P(W > 10) = 5/15 = 1/3.
Lengths are rounded to the nearest cm. Find the probability that the rounding error is more than 0.3 cm in size (either direction). Give your answer as a fraction in its simplest form.
Show answer
2/5. E ~ U[-0.5, 0.5]; |E| > 0.3 has total width 0.2 + 0.2 = 0.4, so 0.4 = 2/5.
Using P(|E| > 0.3) = 0.4 for one rounding error, find the probability that at least 2 of 5 independent rounding errors are more than 0.3 cm in size. Give your answer to 4 decimal places.
Show answer
0.6630. Y ~ B(5, 0.4): 1 - P(Y ≤ 1) = 1 - 0.3370 = 0.6630.
The side of a square, X cm, is modelled as U[1, 4]. Find the expected area of the square, in cm².
Show answer
7 cm². E(X²) = Var(X) + [E(X)]² = 9/12 + 6.25 = 7.
A train is due at 10:00 and arrives at a random time between 9:58 and 10:05. Which model is right for T, the minutes after 10:00 that it arrives?
- U[0, 5]
- U[0, 7]
- U[-2, 5]
- U[-5, 2]
Show answer
U[-2, 5]. Arrival ranges from 2 minutes early (-2) to 5 minutes late (5).

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Last updated 8 November 2026. Spotted a mistake? Tell us.