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The continuous uniform distribution questions
A flat density on an interval, which sounds too simple to be worth a name. It is the model for rounding error and waiting time, and its mean and variance come out of the general formulae in three lines.
6 original questions · 26 marks · the the continuous uniform distribution notes · Further Statistics 2
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Write down the density and the cumulative distribution function of the continuous uniform distribution on the interval a to b.
Worked answer
f(x) = 1/(b − a) for a ≤ x ≤ b and zero elsewhere. F(x) = 0 for x < a, (x − a)/(b − a) for a ≤ x ≤ b, and 1 for x > b. B1 for the density, B1 B1 for the three branches of F. Both branches outside the interval are part of the answer.X follows U(5, 17). Find the mean, the variance and the standard deviation.
Worked answer
Mean = (5 + 17)/2 = 11. Variance = (17 − 5)²/12 = 144/12 = 12. Standard deviation = √12 = 3.46 to three significant figures. B1 for the mean, M1 for the variance formula, A1 for 12 and 3.46. That is about 29% of the range for any uniform distribution, since 1/√12 = 0.289.For the same X, find P(X < 8), P(9 < X < 15) and P(X > 14 given X > 11).
Worked answer
The height is 1/12. P(X < 8) = 3/12 = 0.25. P(9 < X < 15) = 6/12 = 0.5. For the conditional probability, P(X > 14 and X > 11) = P(X > 14) = 3/12, and P(X > 11) = 6/12, so the answer is (3/12)/(6/12) = 0.5. B1 for 0.25, B1 for 0.5, M1 for the conditional form, A1 for 0.5.Readings are rounded to the nearest 10 units. State the distribution of the rounding error, find its standard deviation, and find the probability that it exceeds 4 units in size.
Worked answer
The error is equally likely anywhere in the rounding interval, so it follows U(−5, 5) with mean 0. Variance = 10²/12 = 8.333, so the standard deviation is 2.89. An error above 4 in size covers two stretches of length 1 out of a range of 10, so the probability is 2/10 = 0.2. B1 for U(−5, 5), M1 for the variance formula, A1 for 2.89, M1 for the two stretches of length 1, A1 for 0.2.Buses leave every 15 minutes and a passenger arrives at a random time. Find the expected wait and the probability of waiting more than 10 minutes.
Worked answer
The wait follows U(0, 15), so the expected wait is the midpoint, 7.5 minutes. P(wait > 10) = 5/15 = 1/3. B1 for U(0, 15), B1 for 7.5 minutes, B1 for 1/3. The model assumes arrival independent of the timetable, which fails for a passenger who has checked it.For X following U(a, b), derive the cumulative distribution function, the mean and the variance by integration.
Worked answer
F(x) = ∫ 1/(b − a) dt from a to x = (x − a)/(b − a) for a ≤ x ≤ b. That integral only gives the middle branch, and a cumulative distribution function has to be defined for every real x, so state it in full: F(x) = 0 for x < a, (x − a)/(b − a) for a ≤ x ≤ b, and 1 for x > b. The branches meet, since the middle one is 0 at a and 1 at b, so F is continuous and non-decreasing across the whole line. E(X) = ∫ x/(b − a) dx from a to b = (b² − a²)/[2(b − a)]. Factorising the difference of squares gives (b + a)(b − a)/[2(b − a)] = (a + b)/2, the midpoint, which the symmetry of the rectangle would have predicted.
For the variance, E(X²) = ∫x²/(b − a) dx from a to b = (b³ − a³)/[3(b − a)]. Factorising the difference of cubes gives (b² + ab + a²)/3.
Then Var(X) = E(X²) − [E(X)]² = (b² + ab + a²)/3 − (a + b)²/4. Over a common denominator of 12 that is [4b² + 4ab + 4a² − 3a² − 6ab − 3b²]/12 = (a² − 2ab + b²)/12 = (b − a)²/12.
M1 for integrating the density, A1 for F(x), M1 for ∫x/(b − a) dx, A1 for the mean, M1 for E(X²), A1 for (b² + ab + a²)/3, M1 for E(X²) − [E(X)]², A1 for the printed variance. Both factorisations, of the difference of squares and of the difference of cubes, are where this derivation is usually abandoned. Write them out in full.
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