Edexcel IAL Maths
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The continuous uniform (rectangular) distribution on [a,b]: its pdf and cdf
Deriving the mean (a+b)/2 and variance ((b-a)²)/12, and finding E(g(X)) for a uniform variable
Modelling with the uniform distribution, including rounding errors
The Normal approximation to the binomial, B(n,p)≈N(np,np(1-p)), and when it is valid
The Normal approximation to the Poisson, Po(λ)≈N(λ,λ), for large λ
Continuity corrections for any inequality or single value
Choosing the appropriate approximation, and using one in reverse with a percentage point
For X~U[a,b] the pdf is a rectangle of height 1/(b-a), so every probability is the length of an interval divided by b-a.
E(X)=(a+b)/2, Var(X)=((b-a)²)/12, and F(x)=(x-a)/(b-a) on [a,b].
A value rounded to the nearest k has a rounding error ~U[-k/2,k/2].
B(n,p)≈N(np,np(1-p)) when n is large and p is close to ½; Po(λ)≈N(λ,λ) when λ is large (more than about 10).
Continuity correction: each integer k becomes the interval k-0.5 to k+0.5, so P(X≤38)≈P(Y<38.5) and P(X≥36)≈P(Y>35.5).
Choosing: n large, p small → Poisson; n large, p near ½ → Normal; Poisson with large λ → Normal.
Some quantities are equally likely to be anywhere in a fixed range and cannot fall outside it: the waiting time at a crossing whose signal cycles regularly, the point where a rod snaps, the error left after rounding a reading. The model for all of these is the simplest pdf there is — a flat line.
The continuous uniform (rectangular) distribution
on [a,b] has probability density function
f(x)=1/(b-a) for a≤x≤b, f(x)=0 otherwise.
We write X~U[a,b] (some books write U(a,b)).
The graph of f is a rectangle of width b-a and height 1/(b-a), so its total area is 1, as every pdf must have.
Probabilities are areas of thinner rectangles, so for a≤c<d≤b: P(c<X<d)=(d-c)/(b-a). No integration is needed.
For any continuous variable P(X=c)=0, so < and ≤ give the same probability. (This is not true for the binomial and Poisson distributions later on this page.)
The distribution is symmetrical about its midpoint, so the mean and the median are both (a+b)/2. It has no mode: every value in the range is equally likely.
WORKED EXAMPLE · QUESTION
After a pedestrian presses the button at a crossing, the time T seconds until the green signal appears is modelled by the continuous uniform distribution on [5,35]. (a) Write down the probability density function of T. (b) Find P(T>26). (c) Find E(T) and the standard deviation of T. (d) Find the probability that T is within one standard deviation of its mean.
SOLUTION
(a) Height = 1 ÷ width: f(t)=(1/30) for 5≤t≤35, 0 otherwise
(b) Length of interval × height: P(T>26)=(35-26)×(1/30)= 0.3
(c) Using the formulae (both in the booklet): E(T)=(5+35)/2= 20 s; Var(T)=30²/12=75⇒σ=√75= 8.66 s
(d) The interval 20±8.66 lies inside [5,35], so its probability is its length ÷ 30:
P(20-√75<T<20+√75)=2√75/30=1/√3=
0.577
For any uniform distribution this probability is 1/√3≈0.577, whatever a and b are: the standard deviation is always width ÷√12.
Common error — a pdf without its range
Giving only "f(t)=(1/30)" for a pdf. A pdf is not complete without its range: write "for 5≤t≤35, and 0 otherwise".
Mean and variance of X~U[a,b]
(in the formulae booklet, but the specification also expects you to derive them)
E(X)=(a+b)/2, Var(X)=((b-a)²)/12.
Both come from the general results for a continuous random variable in S2.2: E(X)=∫xf(x) dx and Var(X)=∫x²f(x) dx-[E(X)]². The only algebra you need is two factorisations.
WORKED EXAMPLE · QUESTION
The random variable X has the continuous uniform distribution on [a,b]. Show, by integration, that (a) E(X)=½(a+b) and (b) Var(X)=(1/12)(b-a)².
SOLUTION
(a) Integrate xf(x) over [a,b]:
E(X)=∫ₐᵇx/(b-a) dx=(b²-a²)/(2(b-a))=((b-a)(b+a))/(2(b-a))=
½(a+b)
(b) Find E(X²) in the same way, using b³-a³=(b-a)(b²+ab+a²):
E(X²)=∫ₐᵇx²/(b-a) dx=(b³-a³)/(3(b-a))=(a²+ab+b²)/3.
Subtract the square of the mean over a common denominator of 12:
Var(X)=(a²+ab+b²)/3-((a+b)²)/4=(4a²+4ab+4b²-3a²-6ab-3b²)/12=(a²-2ab+b²)/12=
(1/12)(b-a)²
Part (a) can also be argued from symmetry: the pdf is symmetrical about x=(a+b)/2. But when the question says "by integration", the integral is what is required.
Exam tip — show every factorisation
In a "show that" derivation the answer is printed, so every factorisation must be visible: write b²-a²=(b-a)(b+a) and the expansion over 12 on separate lines, and end with the exact printed form.
The cdf is F(x)=P(X≤x). For U[a,b], integrate the constant pdf from a to x:
Cumulative distribution function of U[a,b]
(not in the formulae booklet — derive it from ∫ₐˣ f(t) dt)
F(x)={ 0, x<a; [2pt] (x-a)/(b-a), a≤x≤b; [4pt] 1, x>b }
All three pieces are part of the answer. The graph is a straight line from (a,0) to (b,1), flat at 0 before and at 1 after.
Quartiles and percentiles: solve F(x)=0.25, 0.5, 0.75,…. The rth percentile is a+r/100(b-a), a fraction of the way along the interval.
If a question gives the mean and variance, use them as two simultaneous equations: a+b=2E(X) and (b-a)²=12 Var(X). Take the positive square root, since b>a.
WORKED EXAMPLE · QUESTION
The continuous random variable X is uniformly distributed on [a,b]. E(X)=7 and Var(X)=3. (a) Find the values of a and b. (b) Specify fully the cumulative distribution function of X. (c) Find the upper quartile of X. (d) Find P(|X-7|<2).
SOLUTION
(a) Two equations: (a+b)/2=7⇒a+b=14, and ((b-a)²)/12=3⇒b-a=6. So a=4, b=10
(b) All three pieces: F(x)=0 for x<4; (x-4)/6 for 4≤x≤10; 1 for x>10
(c) Solve F(q)=0.75: (q-4)/6=0.75⇒ q=8.5
(d) |X-7|<2 means 5<X<9, an interval of length 4: P(5<X<9)=F(9)-F(5)=(4/6)= ⅔
When a measurement is rounded to the nearest k, the error (true value − recorded value) can be anything from -k/2 to k/2, with no reason to favour any part of that range. So the rounding error is modelled by U[-k/2,k/2]:
to the nearest whole number: U[-0.5,0.5]; to the nearest 0.1: U[-0.05,0.05]; to the nearest 5: U[-2.5,2.5];
the mean error is 0 and the variance is k²/12;
if a value is truncated (chopped down) rather than rounded, the error is U[0,k] instead, and its mean is k/2, not 0.
Other natural uniform models are the position of a random break along a rod, the time of an event during a fixed interval, and an angle chosen at random.
Expected values of functions: E(g(X))
E(g(X))=∫g(x)f(x) dx. For linear functions use E(cX+d)=cE(X)+d and Var(cX+d)=c²Var(X). For E(X²) the quickest route is E(X²)=Var(X)+[E(X)]². Example: a disc has radius R cm with R~U[2,4]. Then E(R²)=(4/12)+3²=(28/3), so the expected area is E(πR²)=28π/3≈29.3 cm², a little more than π×3²=9π.
WORKED EXAMPLE · QUESTION
A stopwatch displays times to the nearest 0.1 second. The rounding error E seconds is the true time minus the displayed time. (a) Suggest a distribution for E. (b) Find P(|E|>0.03). (c) Find the standard deviation of E. (d) Find E(E²).
SOLUTION
(a) Rounding to the nearest 0.1 means an error of at most 0.05 either way: E~U[-0.05,0.05]
(b) |E|>0.03 is the two end pieces, −0.05 to −0.03 and 0.03 to 0.05:
P=(0.02+0.02)÷0.1= 0.4
(c) Variance = width²/12: Var(E)=0.1²/12=(1/1200)⇒σ=0.1/√12= 0.0289 s
(d) E(E)=0, so E(E²)=Var(E)+0²= (1/1200)≈0.000833 s²
Common error — an off-centre error interval
Writing U[0,0.1] for the error of a value rounded to the nearest 0.1. Rounding can push a value up or down, so the error interval is centred on zero. U[0,0.1] would describe a value that is always chopped down.
The binomial tables stop at a limited set of values of n and p, and summing dozens of binomial terms by hand is impractical. When the binomial distribution is roughly bell-shaped, a Normal curve with the same mean and variance gives an accurate shortcut. That happens when n is large and p is close to ½. As a working guide, check that np and n(1-p) are both greater than 5.
If p is small (or close to 1), the binomial distribution is skewed and the Normal curve fits badly; with p small and n large, use the Poisson approximation from S2.1 instead.
Normal approximation to the binomial
(not in the formulae booklet: the binomial mean np and variance np(1-p) are given; the approximation and its conditions are not)
X~B(n,p) ⇒ X≈Y, where Y~N(np, np(1-p)),
for n large and p close to ½ (both np and n(1-p) more than about 5). The second parameter is the variance; the standard deviation is √(np(1-p)).
WORKED EXAMPLE · QUESTION
At a garden centre, 35% of customers buy a plant. On one morning 120 customers visit, independently of one another. Using a suitable approximation, find the probability that at most 38 of them buy a plant.
SOLUTION
Define the exact distribution and justify the approximation: X~B(120,0.35); n is large and np=42, n(1-p)=78 are both large, so use a Normal approximation.
Match the mean and variance: Y~N(42,27.3) (variance =120×0.35×0.65=27.3).
Continuity correction — "at most 38" includes the whole bar for 38: P(X≤38)≈P(Y<38.5).
Standardise, rounding z to 2 d.p. for the table:
P(Z<(38.5-42)/√27.3)=P(Z<-0.67)=1-Φ(0.67)=1-0.7486=
0.2514
The exact binomial probability is 0.2533, so the approximation is good to 2 significant figures. A calculator's Normal function (without rounding z) gives 0.2515; either is acceptable.
Exam tip — write the approximating distribution
Write it in full, "Y~N(42,27.3)", before any probability. It shows the method, and it reminds you that 27.3 is the variance: dividing by 27.3 instead of √27.3 is a costly slip.
A binomial or Poisson variable takes only whole-number values; a Normal variable is continuous. Think of each whole number k as a bar from k-0.5 to k+0.5. The continuity correction replaces "these whole numbers" by "the whole width of their bars". Decide which integers are included, then go half a unit beyond the outermost ones:
| Discrete X | Integers included | Continuous Y |
|---|---|---|
| P(X≤38) | …, 37, 38 | P(Y<38.5) |
| P(X<38) | …, 36, 37 | P(Y<37.5) |
| P(X≥56) | 56, 57, … | P(Y>55.5) |
| P(X>56) | 57, 58, … | P(Y>56.5) |
| P(X=50) | 50 only | P(49.5<Y<50.5) |
| P(45≤X<53) | 45, 46, …, 52 | P(44.5<Y<52.5) |
For the continuous Y, < and ≤ are the same, so it does not matter which you write on the right-hand side. Without a continuity correction, P(X=50) would become P(Y=50)=0, which is plainly wrong. The correction is what makes the approximation work.
WORKED EXAMPLE · QUESTION
A quiz has 100 true-or-false questions. A contestant guesses every answer, so the number X of correct answers has distribution B(100,0.5). Use a Normal approximation to find (a) P(X=50), (b) P(X≥56), (c) P(45≤X<53).
SOLUTION
Parameters: np=50, np(1-p)=25⇒Y~N(50,5²).
(a) One integer is a bar of width 1: P(49.5<Y<50.5)=P(-0.1<Z<0.1)=2Φ(0.1)-1=2×0.5398-1= 0.0796
(b) 56 is included, so start at 55.5: P(Y>55.5)=P(Z>1.1)=1-0.8643= 0.1357
(c) Integers 45 to 52: P(44.5<Y<52.5)=P(-1.1<Z<0.5)=0.6915-(1-0.8643)= 0.5558
The exact binomial values are 0.0796, 0.1356 and 0.5557: with p=0.5 the approximation is excellent.
Common error — correcting the wrong way
P(X≥56)→P(Y>56.5) throws away the bar for 56. Always list the integers that are included and move half a unit outwards from them.
The Poisson distribution has mean and variance both equal to λ. As λ grows, its slight right skew fades and it becomes bell-shaped. For large λ (as a guide, λ>10) use N(λ,λ), with a continuity correction as before, since a Poisson variable is also a whole-number count. If the rate is given per unit of time or space, scale λ first (2.5 per minute → 30 per 12 minutes), then approximate.
Normal approximation to the Poisson
(not in the formulae booklet — learn it; the Poisson mean and variance λ are given)
X~Po(λ) ⇒ X≈Y, where Y~N(λ,λ),
valid for large λ. The standard deviation is √λ.
Choosing the approximation
| Exact distribution | Conditions | Approximation |
|---|---|---|
| B(n,p) | n large, p small (np moderate, say up to about 10) | Po(np) — no continuity correction needed, both discrete (S2.1) |
| B(n,p) | n large, p close to ½ | N(np,np(1-p)) with continuity correction |
| Po(λ) | λ large (more than about 10) | N(λ,λ) with continuity correction |
If p is close to 1, count failures instead: the number of failures is B(n,1-p) with a small probability. Use the exact tables whenever n, p or λ are in them: an approximation is only for when the exact route is impractical.
WORKED EXAMPLE · QUESTION
Calls arrive at a help desk at random at a mean rate of 2.5 per minute. (a) Using a suitable approximation, find the probability that more than 35 calls arrive in a 12-minute period. (b) For each situation, name a suitable approximating distribution and use it where asked: (i) 0.75% of laptops sent for repair have a cracked hinge, and 400 laptops are sent; find the probability that at most 2 have a cracked hinge. (ii) 40% of 400 callers ask for a refund.
SOLUTION
(a) Scale the rate, then approximate: X~Po(2.5×12)=Po(30); λ is large, so Y~N(30,30). "More than 35" starts at 36, so the bar edge is 35.5:
P(X>35)≈P(Y>35.5)=P(Z>(35.5-30)/√30)=P(Z>1.00)=1-0.8413=
0.1587
(b)(i) n large, p small → Poisson with λ=np: B(400,0.0075)≈Po(3)⇒P(at most 2)≈ 0.4232 (Poisson table)
(b)(ii) n large, p=0.4 close to ½ → Normal: B(400,0.4)≈N(160,96)
Exact values for comparison: Po(30) gives 0.1574 in (a), and B(400,0.0075) gives 0.4223 in (b)(i).
The two halves of this topic often meet in one question: a uniform model gives a probability p, a count of successes over many repetitions is then binomial with that p, and the Normal approximation answers questions about the count — including a "find the largest value" part that runs the method backwards with a percentage point.
WORKED EXAMPLE · QUESTION
The delay D minutes of a regional train is modelled by the continuous uniform distribution on [-4,16]; a negative delay means the train is early. A journey is late if the delay is more than 10 minutes. Journeys are independent. (a) Find the probability that a journey is late. (b) Find E(D) and the standard deviation of D. (c) Over a season there are 150 journeys. Using a suitable approximation, find the probability that more than 50 of them are late. (d) The operator claims that, with probability at least 0.9, at least m of the 150 journeys will be late. Find the largest value of m for which the claim holds. (e) Give a reason why a Poisson approximation would not be suitable in part (c).
SOLUTION
(a) Length of interval ÷ width: P(D>10)=(16-10)/20= 0.3
(b) E(D)=(-4+16)/2= 6 minutes; Var(D)=20²/12=(100/3)⇒σ= 5.77 minutes
(c) The number late is binomial; approximate it. L~B(150,0.3); n large, np=45⇒Y~N(45,31.5).
P(L>50)≈P(Y>50.5)=P(Z>(50.5-45)/√31.5)=P(Z>0.98)=1-0.8365=
0.1635
(d) With a continuity correction, P(L≥m)≈P(Y>m-0.5)≥0.9. The boundary is in the lower tail, so use z=-1.2816 (10% point):
(m-0.5-45)/√31.5≤-1.2816⇒m≤45.5-1.2816√31.5=45.5-7.193=38.31
so m=38. Check either side: m=38: P(Y>37.5)=Φ(1.34)=0.9099≥0.9 ✓; m=39: P(Y>38.5)=Φ(1.16)=0.8770<0.9 ✗.
(e) p=0.3 is not small (equivalently, the mean 45 and variance 31.5 are not approximately equal, as a Poisson model would need).
The exact binomial answer to (c) is 0.1634, and the exact binomial distribution also gives m=38 in (d).
Exam tip — round in the safe direction
In a "find the largest m" part, keep the inequality sign all the way through and round in the safe direction at the end: the bound 38.31 gives m=38, not 39. Testing the integers either side with the tables is a quick check.
Exam tip 1 — say which distribution, with parameters
Write the exact model first, then the approximation: "X~B(120,0.35), so X≈Y~N(42,27.3)". Parameters with no distribution, or a distribution with no parameters, leave out a step the method needs.
Exam tip 2 — list the integers before correcting
For each probability ask "which whole numbers are included?" and go half a unit outside them. "Fewer than 20" is 0 to 19, so the boundary is 19.5; "at least 20" starts at 20, so the boundary is 19.5 as well, but on the other side.
Exam tip 3 — standard deviation, not variance, in z
N(μ,σ²) is written with the variance. In z=(y-μ)/σ divide by its square root: √27.3, √30, √31.5.
Exam tip 4 — uniform probabilities are lengths
For U[a,b] there is nothing to integrate: P= (length of the interval inside [a,b]) ÷ (b-a). Clip the interval to [a,b] first — P(X>12) for U[4,10] is 0.
Exam tip 5 — justify the approximation in the question's terms
"n is large and p is close to 0.5" or "λ is large" is the reasoning required; a general remark such as "it is easier" is not. When choosing between Poisson and Normal for a binomial, the size of p decides.
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| The variance of U[a,b] is (b-a)/12. | Variance has squared units; the width must be squared. | Var(X)=((b-a)²)/12; standard deviation =(b-a)/√12. |
| The rounding error for "nearest 0.1" is U[0,0.1]. | Rounding can move a value up or down. | The error is U[-0.05,0.05], mean 0. |
| A continuity correction is needed for the uniform distribution. | The uniform distribution is already continuous. | Correct only when a discrete count (binomial, Poisson) is replaced by a Normal variable. |
| P(X<20) and P(X≤20) have the same Normal approximation. | For a discrete X they include different integers. | P(X<20)≈P(Y<19.5); P(X≤20)≈P(Y<20.5). |
| Any binomial with large n can be approximated by a Normal distribution. | With small p the distribution is skewed even for large n. | Normal needs p near ½; for small p use Po(np). |
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Po(λ)≈N(λ,√λ). | The second parameter of N is the variance, and the Poisson variance is λ. | Po(λ)≈N(λ,λ); standardise with √λ. |
S1.6 The Normal distribution
Every approximation on this page ends with standardising and the Φ(z) or percentage-point tables, exactly as in S1; only the continuity correction is new.
S2.1 and S2.2
The approximations replace the exact binomial and Poisson models from S2.1, and the uniform distribution is the simplest case of the general pdf, cdf, E(X) and Var(X) methods from S2.2.
S2.4 Hypothesis tests
Tests on a binomial proportion or a Poisson rate with large numbers use these same Normal approximations to find critical regions. In IAL Physics, a reading rounded to the smallest scale division has an error spread evenly over ± half a division, and large radioactive counts behave approximately Normally with standard deviation √(count).
| Formula | Use | In the formulae list? |
|---|---|---|
| f(x)=1/(b-a), a≤x≤b | pdf of U[a,b] | Yes |
| E(X)=(a+b)/2, Var(X)=((b-a)²)/12 | mean and variance of U[a,b] (derivation also required) | Yes |
| F(x)=(x-a)/(b-a), a≤x≤b (0 below, 1 above) | cdf of U[a,b]; quartiles and percentiles | No – derive it |
| E(g(X))=∫g(x)f(x) dx; E(X²)=Var(X)+[E(X)]² | expected value of a function of X | Yes |
| rounding to nearest k: error ~U[-k/2,k/2] | modelling rounding errors | No – derive it |
| B(n,p)≈N(np,np(1-p)) | n large, p close to ½ | No – learn it |
| Po(λ)≈N(λ,λ) | λ large | No – learn it |
| P(X≤k)≈P(Y<k+0.5), P(X≥k)≈P(Y>k-0.5) | continuity correction | No – learn it |
| z=(y-μ)/σ | standardising before using Φ(z) | No – learn it |
1. X~U[-2,6]. Find P(X<0) and Var(X).
2. X~B(80,0.45). Write P(X<30) in terms of the approximating Normal variable Y.
3. Which approximation would you use for B(500,0.004), and why?
4. Why is a Normal approximation unsuitable for B(30,0.05)?
5. X~Po(64). Use a Normal approximation to find P(X≥70).
6. A length is recorded to the nearest 5 mm. Find the standard deviation of the rounding error.
1. P(X<0)=(2/8)=0.25. Var(X)=8²/12=(16/3).
2. "Fewer than 30" means 29 or fewer, so P(Y<29.5), with Y~N(36,19.8).
3. Po(2): n is large and p is small, and np=2.
4. np=1.5 is small: the distribution is strongly skewed, and a Normal curve would give noticeable probability to negative values. (A Poisson approximation Po(1.5) could be used, or the exact binomial.)
5. Y~N(64,64); P(Y>69.5)=P(Z>5.5/8)=P(Z>0.69)=1-0.7549=0.2451.
6. The error is U[-2.5,2.5], width 5, so σ=5/√12=1.44 mm (3 s.f.).
Check understanding
The answer and each step appear one click at a time.
A computer program generates a random number R from the continuous uniform distribution on [−3, 5]. Successive numbers are independent.
(a) Specify fully the cumulative distribution function of R. [3]
(b) Find the 20th percentile of R. [1]
(c) Find E(4R − 3) and Var(4R − 3). [3]
(d) The program generates three numbers. Find the probability that all three are positive. [2]
Mark scheme click to reveal each step
F(r) = (r + 3)/8 for −3 ≤ r ≤ 5
F(r) = 0 for r < −3
F(r) = 1 for r > 5
(r + 3)/8 = 0.2 ⟹ r = −1.4
E(R) = 1, so E(4R − 3) = 1
Var(R) = 8²/12 = 16/3
Var(4R − 3) = 16 × 16/3 = 256/3 (≈ 85.3)
P(R > 0) = 5/8, cubed for three independent numbers
The time T seconds that a laser printer takes to warm up has the cumulative distribution function F(t) shown in Figure 1. The graph consists of straight-line sections.
(a) Explain how Figure 1 shows that T has a continuous uniform distribution, and state the interval on which it is defined. [2]
(b) Specify fully the probability density function of T. [1]
(c) Find P(2 < T < 3.2). [2]
(d) Find the value of t such that P(T > t) = 0.1. [1]
Mark scheme click to reveal each step
F rises at a constant rate (a straight line) between its two corners, so the pdf, its gradient, is constant there
T ~ U[1.5, 4]
f(t) = 0.4 for 1.5 ≤ t ≤ 4, and f(t) = 0 otherwise
(3.2 − 2)/2.5 or F(3.2) − F(2) = 0.68 − 0.2
= 0.48
(t − 1.5)/2.5 = 0.9 ⟹ t = 3.75
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