Questions on the continuous uniform distribution

Question 1

Write down the density and the cumulative distribution function of the continuous uniform distribution on the interval a to b.

Accepted 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.

Question 2

X follows U(5, 17). Find the mean, the variance and the standard deviation.

Accepted 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.

Question 3

For the same X, write down F(x) for 5 ≤ x ≤ 17, and write down the value of F(11).

Accepted Answer

F(x) = (x − 5)/12 on 5 ≤ x ≤ 17, with F = 0 below 5 and F = 1 above 17. B1.
F(11) = 6/12 = 0.5. B1.
The value 11 is the midpoint of the interval and also the mean, and for a uniform distribution it is the median as well, which is what F(11) = 0.5 records.

Question 4

For the same X, find the interquartile range and the 90th percentile.

Accepted Answer

The quartiles of a uniform distribution sit a quarter and three quarters of the way along the interval, so Q₁ = 5 + 0.25(12) = 8 and Q₃ = 5 + 0.75(12) = 14. M1 for the quartile positions.
The interquartile range is 14 − 8 = 6, exactly half the width of the interval, which is true of every continuous uniform distribution. A1.
The 90th percentile solves F(x) = 0.9, giving x = 5 + 0.9(12) = 15.8. A1.
Each of these is a fraction of the width added to the lower end. Taking the same fraction of the upper end instead gives 0.25 × 17 = 4.25, which is not even inside the interval.

Question 5

For the same X, find P(X < 8), P(9 < X < 15) and P(X > 14 given X > 11).

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

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.

Accepted 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.

Question 8

The time X minutes that a machine takes to complete one cycle is modelled by a continuous uniform distribution on the interval 2 to 10. Write down F(x) for 2 ≤ x ≤ 10, and use it to find P(1 < X < 5) and P(X > 12).

Accepted Answer

F(x) = (x − 2)/8 on that interval.
P(1 < X < 5) = F(5) − F(1) = 3/8 − 0 = 0.375. The interval asked about starts at 1, but the distribution puts no probability below 2, so the length that counts is 5 − 2 = 3 and not 5 − 1 = 4.
P(X > 12) = 0. The support ends at 10, so F(x) has already reached 1 there and there is nothing left above it.
B1 for F(x), M1 for using only the part of the interval inside the support, A1 for 0.375, B1 for 0.

Question 9

The random variable X follows U(0, 1). Find the distribution function of Y = 5X + 2, and hence name the distribution of Y.

Accepted Answer

X runs from 0 to 1 and Y = 5X + 2 is increasing, so Y runs from 2 to 7. B1 for the range.
For 2 ≤ y ≤ 7, the event Y ≤ y says the same thing as X ≤ (y − 2)/5, and F_X(x) = x on 0 ≤ x ≤ 1, so
F_Y(y) = P(X ≤ (y − 2)/5) = (y − 2)/5, with 0 below 2 and 1 above 7. M1 for rewriting the event about X, A1 for F_Y with its branches.
That is the distribution function of a continuous uniform distribution on 2 to 7, so Y is U(2, 7). A1.
Differentiating confirms it: f_Y(y) = 1/5, a constant across an interval of width 5, whose area is 1. A linear change of variable turns one uniform distribution into another, which is why U(0, 1) is enough to generate any of them.

Question 10

The random variable X follows a continuous uniform distribution on the interval a to b. Given that P(X < 10) = 0.25 and P(X < 22) = 0.85, find a and b.

Accepted Answer

With F(x) = (x − a)/(b − a) on the interval, the two statements give
(10 − a)/(b − a) = 0.25 and (22 − a)/(b − a) = 0.85. M1 for both equations.
Subtracting the first from the second removes a entirely: 12/(b − a) = 0.60. M1 for the subtraction.
So b − a = 12/0.6 = 20. A1.
Then 10 − a = 0.25 × 20 = 5, so a = 5 and b = 25. A1.
Check both givens: (10 − 5)/20 = 0.25 and (22 − 5)/20 = 0.85, and both 10 and 22 lie inside the interval, as they must for those two equations to have applied at all.

Question 11

A person arrives at a meeting point at a time X minutes after 1 pm, where X is modelled by U(0, 60). Find P(X > 45), find the value of t for which P(X > t) = 0.35, and find E(X²).

Accepted Answer

The height of the density is 1/60, so probabilities read off as lengths divided by 60.
P(X > 45) = 15/60 = 0.25. B1.
P(X > t) = (60 − t)/60, so (60 − t)/60 = 0.35 gives 60 − t = 21 and t = 39. M1 for the equation, A1 for 39.
For the second moment, use E(X²) = Var(X) + [E(X)]². Here Var(X) = 60²/12 = 300 and E(X) = 30, so E(X²) = 300 + 900 = 1200. A1.
Integrating ∫x²/60 dx from 0 to 60 gives [x³/180] = 216 000/180 = 1200 as well, which is the check worth making once. Note that E(X²) = 1200 is not the square of E(X) = 30.

Question 12

For X following U(a, b), derive the cumulative distribution function, the mean and the variance by integration.

Accepted 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.

Question 13

X follows a continuous uniform distribution on the interval a to b, where a < b. Given that E(X) = 8 and Var(X) = 12, find a and b, and hence find the interquartile range of X and P(X > 11).

Accepted Answer

The mean gives (a + b)/2 = 8, so a + b = 16. The variance gives (b − a)²/12 = 12, so (b − a)² = 144 and b − a = 12, taking the positive root because b > a. Adding the two equations gives 2b = 28, so b = 14 and a = 2.
The quartiles of a uniform distribution sit a quarter and three quarters of the way along the interval, so Q₁ = 2 + 12/4 = 5 and Q₃ = 2 + 3(12)/4 = 11. The interquartile range is 11 − 5 = 6, which is half the range for any continuous uniform distribution.
P(X > 11) = (14 − 11)/12 = 0.25, and that agrees with the quartiles rather than being a separate result: 11 is Q₃, so a quarter of the distribution lies above it.
M1 for the equation from the mean, M1 for the equation from the variance, A1 for b − a = 12, A1 for a = 2 and b = 14, M1 for the quartile positions, A1 for the interquartile range 6, B1 for 0.25.
The root b − a = −12 is not a second answer. It is the same pair of numbers with the labels exchanged, and it contradicts the given condition a < b.

Question 14

A rod of length 12 cm is cut at a point chosen at random along its length, so that the distance X cm from one end to the cut follows U(0, 12). Find the probability that the longer piece is at least twice the length of the shorter, find the probability that the two pieces differ in length by less than 3 cm, and find the value of k for which the shorter piece is less than k cm long half of the time.

Accepted Answer

The two pieces are X and 12 − X, and every part of this question is the same manoeuvre: turn the statement about the pieces into a set of values of X, then read the probability off as a length divided by 12.
If the shorter piece has length s, the longer has 12 − s, and 12 − s ≥ 2s exactly when s ≤ 4. So the condition is X ≤ 4 or X ≥ 8, two stretches of length 4, and the probability is 8/12 = 2/3.
The difference in length is |X − (12 − X)| = |2X − 12|, which is less than 3 when |X − 6| < 1.5, that is 4.5 < X < 7.5. That is one stretch of length 3, so the probability is 3/12 = 0.25.
For any k below 6, the shorter piece is less than k when X < k or X > 12 − k. The two stretches have total length 2k, so P(shorter < k) = 2k/12 = k/6. Setting k/6 = 0.5 gives k = 3.
M1 for translating the condition onto the shorter piece, A1 for X ≤ 4 or X ≥ 8, A1 for 2/3, M1 for the inequality on |2X − 12|, A1 for 0.25, M1 for the total length 2k, A1 for k = 3.
Since P(shorter < k) = k/6 for every k between 0 and 6, the length of the shorter piece is itself uniform on 0 to 6, and 3 is its median as well as its mean.

Question 15

The random variable X follows U(0, 6), and Y = X(6 − X). Find the greatest value Y can take, find P(Y > 8), and find E(Y).

Accepted Answer

The greatest value. Y = 6X − X² = 9 − (X − 3)², which is largest when X = 3. M1 for completing the square or for differentiating. So the greatest value of Y is 9, and X = 3 lies inside 0 to 6, so it is reached. A1. The probability. Y > 8 means X(6 − X) > 8, that is X² − 6X + 8 < 0, which factorises to (X − 2)(X − 4) < 0. M1 for the inequality. A negative product needs X between the roots, so the condition is 2 < X < 4. A1. That is one stretch of length 2 out of a range of 6, so P(Y > 8) = 2/6 = 1/3. A1. The mean. E(Y) = E(6X − X²) = 6E(X) − E(X²). For U(0, 6), E(X) = 3 and Var(X) = 36/12 = 3, so E(X²) = Var(X) + [E(X)]² = 3 + 9 = 12. Hence E(Y) = 18 − 12 = 6. B1. Y is not uniform, so its mean is not the midpoint of the values it takes: the mean 6 sits well below the greatest value 9, because X spends most of its range away from 3. Substituting E(X) = 3 into Y would give 9, the largest value rather than the mean, and E(X²) is 12 rather than 3² = 9.