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IAL Maths (Edexcel) · S2 · Topic S2.3A2≈ 35 min read

S2.3 Continuous uniform distribution and Normal approximations

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A2 content (Statistics 2, specification sections 3.1–3.2); it builds on the Normal tables from S1 and on S2.1 and S2.2.

What this page covers

7 points
  • The continuous uniform (rectangular) distribution on \([a,b]\): its pdf and cdf
  • Deriving the mean \(\frac{a+b}{2}\) and variance \(\frac{(b-a)^2}{12}\), and finding \(E(g(X))\) for a uniform variable
  • Modelling with the uniform distribution, including rounding errors
  • The Normal approximation to the binomial, \(B(n,p)\approx N(np,np(1-p))\), and when it is valid
  • The Normal approximation to the Poisson, \(Po(\lambda)\approx N(\lambda,\lambda)\), for large \(\lambda\)
  • Continuity corrections for any inequality or single value
  • Choosing the appropriate approximation, and using one in reverse with a percentage point

In one minute

6 points
  • For \(X\sim U[a,b]\) the pdf is a rectangle of height \(\frac{1}{b-a}\), so every probability is the length of an interval divided by \(b-a\).
  • \(E(X)=\frac{a+b}{2}\), \(\text{Var}(X)=\frac{(b-a)^2}{12}\), and \(F(x)=\frac{x-a}{b-a}\) on \([a,b]\).
  • A value rounded to the nearest \(k\) has a rounding error \(\sim U\left[-\frac{k}{2},\frac{k}{2}\right]\).
  • \(B(n,p)\approx N(np,np(1-p))\) when \(n\) is large and \(p\) is close to \(\frac12\); \(Po(\lambda)\approx N(\lambda,\lambda)\) when \(\lambda\) is large (more than about 10).
  • Continuity correction: each integer \(k\) becomes the interval \(k-0.5\) to \(k+0.5\), so \(P(X\le38)\approx P(Y\lt38.5)\) and \(P(X\ge36)\approx P(Y\gt35.5)\).
  • Choosing: \(n\) large, \(p\) small → Poisson; \(n\) large, \(p\) near \(\frac12\) → Normal; Poisson with large \(\lambda\) → Normal.

Contents

14 sections
  1. What is the continuous uniform distribution?
  2. Where do the mean and variance of U[a, b] come from?
  3. How do you use the cumulative distribution function?
  4. How does the uniform distribution model rounding errors?
  5. When can a Normal distribution approximate a binomial distribution?
  6. What is a continuity correction, and how do you apply it?
  7. When can a Normal distribution approximate a Poisson distribution, and which approximation should you choose?
  8. How does a full exam question fit together?
  9. Exam tips for S2.3
  10. Common misconceptions
  11. How this topic connects
  12. Key formulae
  13. Check your understanding
  14. Specification checklist

What is the continuous uniform distribution?

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 \[f(x)=\frac{1}{b-a}\ \text{ for } a\le x\le b,\qquad f(x)=0\ \text{ otherwise.}\] We write \(X\sim U[a,b]\) (some books write \(U(a,b)\)).
  • The graph of \(f\) is a rectangle of width \(b-a\) and height \(\frac{1}{b-a}\), so its total area is 1, as every pdf must have.
  • Probabilities are areas of thinner rectangles, so for \(a\le c\lt d\le b\): \(P(c\lt X\lt d)=\dfrac{d-c}{b-a}\). No integration is needed.
  • For any continuous variable \(P(X=c)=0\), so \(\lt\) and \(\le\) 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 \(\frac{a+b}{2}\). It has no mode: every value in the range is equally likely.
05101520253035401/30t (s)f(t)0.3height 1/30P(T > 26)
The pdf of \(T\sim U[5,35]\) is a rectangle of height \(\frac{1}{30}\). The red part, from 26 to 35, has area \(9\times\frac{1}{30}=0.3=P(T\gt26)\).
Worked example 1 Easy · Write down, find

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 \(P(T\gt26)\). (c) Find \(E(T)\) and the standard deviation of \(T\). (d) Find the probability that \(T\) is within one standard deviation of its mean.

(a) Height = 1 ÷ width: \(f(t)=\frac{1}{30}\) for \(5\le t\le35\), 0 otherwise

(b) Length of interval × height: \(P(T\gt26)=(35-26)\times\frac{1}{30}=\) 0.3

(c) Using the formulae (both in the booklet): \(E(T)=\frac{5+35}{2}=\) 20 s; \(\text{Var}(T)=\frac{30^2}{12}=75\Rightarrow\sigma=\sqrt{75}=\) 8.66 s

(d) The interval \(20\pm8.66\) lies inside \([5,35]\), so its probability is its length ÷ 30: \[P(20-\sqrt{75}\lt T\lt20+\sqrt{75})=\frac{2\sqrt{75}}{30}=\frac{1}{\sqrt3}=\] 0.577

For any uniform distribution this probability is \(\frac{1}{\sqrt3}\approx0.577\), whatever \(a\) and \(b\) are: the standard deviation is always width \(\div\sqrt{12}\).

Common error — a pdf without its range

Giving only "\(f(t)=\frac{1}{30}\)" for a pdf. A pdf is not complete without its range: write "for \(5\le t\le35\), and 0 otherwise".

Where do the mean and variance of U[a, b] come from?

Mean and variance of \(X\sim U[a,b]\) (in the formulae booklet, but the specification also expects you to derive them) \[E(X)=\frac{a+b}{2},\qquad \text{Var}(X)=\frac{(b-a)^2}{12}.\]

Both come from the general results for a continuous random variable in S2.2: \(E(X)=\int xf(x)\,dx\) and \(\text{Var}(X)=\int x^2f(x)\,dx-[E(X)]^2\). The only algebra you need is two factorisations.

Worked example 2 Medium · Show that

The random variable \(X\) has the continuous uniform distribution on \([a,b]\). Show, by integration, that (a) \(E(X)=\frac12(a+b)\) and (b) \(\text{Var}(X)=\frac{1}{12}(b-a)^2\).

(a) Integrate \(xf(x)\) over \([a,b]\): \[E(X)=\int_a^b\frac{x}{b-a}\,dx=\frac{b^2-a^2}{2(b-a)}=\frac{(b-a)(b+a)}{2(b-a)}=\] \(\frac12(a+b)\)

(b) Find \(E(X^2)\) in the same way, using \(b^3-a^3=(b-a)(b^2+ab+a^2)\): \[E(X^2)=\int_a^b\frac{x^2}{b-a}\,dx=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3}.\] Subtract the square of the mean over a common denominator of 12: \[\text{Var}(X)=\frac{a^2+ab+b^2}{3}-\frac{(a+b)^2}{4}=\frac{4a^2+4ab+4b^2-3a^2-6ab-3b^2}{12}=\frac{a^2-2ab+b^2}{12}=\] \(\frac{1}{12}(b-a)^2\)

Part (a) can also be argued from symmetry: the pdf is symmetrical about \(x=\frac{a+b}{2}\). 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 \(b^2-a^2=(b-a)(b+a)\) and the expansion over 12 on separate lines, and end with the exact printed form.

How do you use the cumulative distribution function?

The cdf is \(F(x)=P(X\le x)\). 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 \(\int_a^x f(t)\,dt\)) \[F(x)=\begin{cases}0 & x\lt a\\[2pt] \dfrac{x-a}{b-a} & a\le x\le b\\[4pt] 1 & x\gt b\end{cases}\] 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 \(F(x)=0.25,\ 0.5,\ 0.75,\ldots\). The \(r\)th percentile is \(a+\frac{r}{100}(b-a)\), 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 \((b-a)^2=12\,\text{Var}(X)\). Take the positive square root, since \(b\gt a\).
2345678910111200.250.50.751xF(x)median 7Q₃ = 8.5F(x) = (x − 4)/6
The cdf of \(X\sim U[4,10]\): zero before 4, a straight line of gradient \(\frac16\) up to \((10,1)\), then constant at 1. Reading across from 0.5 and 0.75 gives the median 7 and the upper quartile 8.5.
Worked example 3 Medium · Find, specify

The continuous random variable \(X\) is uniformly distributed on \([a,b]\). \(E(X)=7\) and \(\text{Var}(X)=3\). (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 \(P(|X-7|\lt2)\).

(a) Two equations: \(\frac{a+b}{2}=7\Rightarrow a+b=14\), and \(\frac{(b-a)^2}{12}=3\Rightarrow b-a=6\). So \(a=4\), \(b=10\)

(b) All three pieces: \(F(x)=0\) for \(x\lt4\); \(\frac{x-4}{6}\) for \(4\le x\le10\); 1 for \(x\gt10\)

(c) Solve \(F(q)=0.75\): \(\frac{q-4}{6}=0.75\Rightarrow\) \(q=8.5\)

(d) \(|X-7|\lt2\) means \(5\lt X\lt9\), an interval of length 4: \(P(5\lt X\lt9)=F(9)-F(5)=\frac46=\) \(\frac23\)

234567891011121/6xf(x)|x − 7| < 24/6 = 2/3
Part (d): the pdf of \(U[4,10]\). The event \(|X-7|\lt2\) is the strip from 5 to 9, which is \(\frac46\) of the rectangle.

How does the uniform distribution model rounding errors?

When a measurement is rounded to the nearest \(k\), the error (true value − recorded value) can be anything from \(-\frac{k}{2}\) to \(\frac{k}{2}\), with no reason to favour any part of that range. So the rounding error is modelled by \(U\left[-\frac{k}{2},\frac{k}{2}\right]\):

  • 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 \(\frac{k^2}{12}\);
  • if a value is truncated (chopped down) rather than rounded, the error is \(U[0,k]\) instead, and its mean is \(\frac{k}{2}\), 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))

\(E(g(X))=\int g(x)f(x)\,dx\). For linear functions use \(E(cX+d)=cE(X)+d\) and \(\text{Var}(cX+d)=c^2\text{Var}(X)\). For \(E(X^2)\) the quickest route is \(E(X^2)=\text{Var}(X)+[E(X)]^2\). Example: a disc has radius \(R\) cm with \(R\sim U[2,4]\). Then \(E(R^2)=\frac{4}{12}+3^2=\frac{28}{3}\), so the expected area is \(E(\pi R^2)=\frac{28\pi}{3}\approx29.3\) cm², a little more than \(\pi\times3^2=9\pi\).

Worked example 4 Medium · Suggest, find

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 \(P(|E|\gt0.03)\). (c) Find the standard deviation of \(E\). (d) Find \(E(E^2)\).

−0.05−0.0300.030.05510error e (s)f(e)0.20.2height 1/0.1 = 10
The rounding error is equally likely to be anywhere between −0.05 and 0.05. The two red end strips are the event \(|E|\gt0.03\).

(a) Rounding to the nearest 0.1 means an error of at most 0.05 either way: \(E\sim U[-0.05,0.05]\)

(b) \(|E|\gt0.03\) is the two end pieces, −0.05 to −0.03 and 0.03 to 0.05: \(P=(0.02+0.02)\div0.1=\) 0.4

(c) Variance = width²/12: \(\text{Var}(E)=\frac{0.1^2}{12}=\frac{1}{1200}\Rightarrow\sigma=\frac{0.1}{\sqrt{12}}=\) 0.0289 s

(d) \(E(E)=0\), so \(E(E^2)=\text{Var}(E)+0^2=\) \(\frac{1}{1200}\approx0.000833\) 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.

When can a Normal distribution approximate a binomial distribution?

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 \(\frac12\). 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) \[X\sim B(n,p)\ \Rightarrow\ X\approx Y,\ \text{ where } Y\sim N(np,\ np(1-p)),\] for \(n\) large and \(p\) close to \(\frac12\) (both \(np\) and \(n(1-p)\) more than about 5). The second parameter is the variance; the standard deviation is \(\sqrt{np(1-p)}\).
−3−2−1012345678910111200.10.20.3xN(2, 1.8)B(20, 0.1): skewed
Same \(n=20\), \(p=0.1\): the binomial bars are skewed and the matching Normal curve misses them — it even gives probability to negative values, left of the black line.
2345678910111213141516171800.050.10.150.2xN(10, 5)B(20, 0.5): symmetrical
Same \(n=20\), \(p=0.5\): the bars are symmetrical and the curve with the same mean and variance follows them closely.
Worked example 5 Medium · Find

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.

263034384246505458number buying a plantN(42, 27.3)38.5
\(X\sim B(120,0.35)\) drawn as bars of width 1 with \(Y\sim N(42,27.3)\) over them. The bar for 38 runs from 37.5 to 38.5, so \(P(X\le38)\) is approximated by the area under the curve to the left of 38.5.

Define the exact distribution and justify the approximation: \(X\sim B(120,0.35)\); \(n\) is large and \(np=42\), \(n(1-p)=78\) are both large, so use a Normal approximation.

Match the mean and variance: \(Y\sim N(42,27.3)\) (variance \(=120\times0.35\times0.65=27.3\)).

Continuity correction — "at most 38" includes the whole bar for 38: \(P(X\le38)\approx P(Y\lt38.5)\).

Standardise, rounding \(z\) to 2 d.p. for the table: \[P\!\left(Z\lt\frac{38.5-42}{\sqrt{27.3}}\right)=P(Z\lt-0.67)=1-\Phi(0.67)=1-0.7486=\] 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, "\(Y\sim N(42,27.3)\)", before any probability. It shows the method, and it reminds you that 27.3 is the variance: dividing by 27.3 instead of \(\sqrt{27.3}\) is a costly slip.

What is a continuity correction, and how do you apply it?

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 includedContinuous \(Y\)
\(P(X\le38)\)…, 37, 38\(P(Y\lt38.5)\)
\(P(X\lt38)\)…, 36, 37\(P(Y\lt37.5)\)
\(P(X\ge56)\)56, 57, …\(P(Y\gt55.5)\)
\(P(X\gt56)\)57, 58, …\(P(Y\gt56.5)\)
\(P(X=50)\)50 only\(P(49.5\lt Y\lt50.5)\)
\(P(45\le X\lt53)\)45, 46, …, 52\(P(44.5\lt Y\lt52.5)\)

For the continuous \(Y\), \(\lt\) and \(\le\) 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.

4045505560x44.552.5
Close-up of \(B(100,0.5)\) with \(N(50,25)\). The shaded bars are 45, 46, …, 52: the left edge of the first is 44.5 and the right edge of the last is 52.5, so \(P(45\le X\lt53)\approx P(44.5\lt Y\lt52.5)\).
Worked example 6 Medium · Find

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) \(P(X\ge56)\), (c) \(P(45\le X\lt53)\).

Parameters: \(np=50\), \(np(1-p)=25\Rightarrow Y\sim N(50,5^2)\).

(a) One integer is a bar of width 1: \(P(49.5\lt Y\lt50.5)=P(-0.1\lt Z\lt0.1)=2\Phi(0.1)-1=2\times0.5398-1=\) 0.0796

(b) 56 is included, so start at 55.5: \(P(Y\gt55.5)=P(Z\gt1.1)=1-0.8643=\) 0.1357

(c) Integers 45 to 52: \(P(44.5\lt Y\lt52.5)=P(-1.1\lt Z\lt0.5)=0.6915-(1-0.8643)=\) 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

\(P(X\ge56)\to P(Y\gt56.5)\) throws away the bar for 56. Always list the integers that are included and move half a unit outwards from them.

When can a Normal distribution approximate a Poisson distribution, and which approximation should you choose?

The Poisson distribution has mean and variance both equal to \(\lambda\). As \(\lambda\) grows, its slight right skew fades and it becomes bell-shaped. For large \(\lambda\) (as a guide, \(\lambda\gt10\)) use \(N(\lambda,\lambda)\), 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 \(\lambda\) 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 \(\lambda\) are given) \[X\sim Po(\lambda)\ \Rightarrow\ X\approx Y,\ \text{ where } Y\sim N(\lambda,\lambda),\] valid for large \(\lambda\). The standard deviation is \(\sqrt\lambda\).
15202530354045calls in 12 minutesN(30, 30)35.5≈ 0.1587
\(X\sim Po(30)\) with \(Y\sim N(30,30)\). The bars for 36 and above start at 35.5, so \(P(X\gt35)\approx P(Y\gt35.5)\).

Choosing the approximation

Exact distributionConditionsApproximation
\(B(n,p)\)\(n\) 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 \(\frac12\)\(N(np,np(1-p))\) with continuity correction
\(Po(\lambda)\)\(\lambda\) large (more than about 10)\(N(\lambda,\lambda)\) 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 \(\lambda\) are in them: an approximation is only for when the exact route is impractical.

Worked example 7 Medium · Find, name

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.

(a) Scale the rate, then approximate: \(X\sim Po(2.5\times12)=Po(30)\); \(\lambda\) is large, so \(Y\sim N(30,30)\). "More than 35" starts at 36, so the bar edge is 35.5: \[P(X\gt35)\approx P(Y\gt35.5)=P\!\left(Z\gt\frac{35.5-30}{\sqrt{30}}\right)=P(Z\gt1.00)=1-0.8413=\] 0.1587

(b)(i) \(n\) large, \(p\) small → Poisson with \(\lambda=np\): \(B(400,0.0075)\approx Po(3)\Rightarrow P(\text{at most }2)\approx\) 0.4232 (Poisson table)

(b)(ii) \(n\) large, \(p=0.4\) close to \(\frac12\) → Normal: \(B(400,0.4)\approx N(160,96)\)

Exact values for comparison: \(Po(30)\) gives 0.1574 in (a), and \(B(400,0.0075)\) gives 0.4223 in (b)(i).

How does a full exam question fit together?

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 8 Hard · Find, give a reason

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).

(a) Length of interval ÷ width: \(P(D\gt10)=\frac{16-10}{20}=\) 0.3

(b) \(E(D)=\frac{-4+16}{2}=\) 6 minutes; \(\text{Var}(D)=\frac{20^2}{12}=\frac{100}{3}\Rightarrow\sigma=\) 5.77 minutes

(c) The number late is binomial; approximate it. \(L\sim B(150,0.3)\); \(n\) large, \(np=45\Rightarrow Y\sim N(45,31.5)\). \[P(L\gt50)\approx P(Y\gt50.5)=P\!\left(Z\gt\frac{50.5-45}{\sqrt{31.5}}\right)=P(Z\gt0.98)=1-0.8365=\] 0.1635

(d) With a continuity correction, \(P(L\ge m)\approx P(Y\gt m-0.5)\ge0.9\). The boundary is in the lower tail, so use \(z=-1.2816\) (10% point): \[\frac{m-0.5-45}{\sqrt{31.5}}\le-1.2816\Rightarrow m\le45.5-1.2816\sqrt{31.5}=45.5-7.193=38.31\] so \(m=38\). Check either side: \(m=38\): \(P(Y\gt37.5)=\Phi(1.34)=0.9099\ge0.9\) ✓; \(m=39\): \(P(Y\gt38.5)=\Phi(1.16)=0.8770\lt0.9\) ✗.

(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).

30354045505560number of late journeys37.5 (m = 38)38.5 (m = 39)50.50.1635
\(Y\sim N(45,31.5)\) for the number of late journeys. The amber tail beyond 50.5 is about 0.1635. For part (d), the area to the right of 37.5 (\(m=38\)) is about 0.91, at least 0.9; for \(m=39\) the boundary moves to 38.5 and the area drops to about 0.877.
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 tips for S2.3

Exam tip 1 — say which distribution, with parameters

Write the exact model first, then the approximation: "\(X\sim B(120,0.35)\), so \(X\approx Y\sim N(42,27.3)\)". 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

\(N(\mu,\sigma^2)\) is written with the variance. In \(z=\frac{y-\mu}{\sigma}\) divide by its square root: \(\sqrt{27.3}\), \(\sqrt{30}\), \(\sqrt{31.5}\).

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 — \(P(X\gt12)\) 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 "\(\lambda\) 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.

Common misconceptions

MisconceptionWhy it is wrongCorrect idea
The variance of \(U[a,b]\) is \(\frac{b-a}{12}\).Variance has squared units; the width must be squared.\(\text{Var}(X)=\frac{(b-a)^2}{12}\); standard deviation \(=\frac{b-a}{\sqrt{12}}\).
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.
\(P(X\lt20)\) and \(P(X\le20)\) have the same Normal approximation.For a discrete \(X\) they include different integers.\(P(X\lt20)\approx P(Y\lt19.5)\); \(P(X\le20)\approx P(Y\lt20.5)\).
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 \(\frac12\); for small \(p\) use \(Po(np)\).
\(Po(\lambda)\approx N(\lambda,\sqrt\lambda)\).The second parameter of \(N\) is the variance, and the Poisson variance is \(\lambda\).\(Po(\lambda)\approx N(\lambda,\lambda)\); standardise with \(\sqrt\lambda\).

How this topic connects

S1.6 The Normal distribution

Every approximation on this page ends with standardising and the \(\Phi(z)\) 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 \(\text{Var}(X)\) 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 \(\sqrt{\text{count}}\).

Key formulae

FormulaUseIn the formulae list?
\(f(x)=\frac{1}{b-a}\), \(a\le x\le b\)pdf of \(U[a,b]\)Yes
\(E(X)=\frac{a+b}{2}\), \(\text{Var}(X)=\frac{(b-a)^2}{12}\)mean and variance of \(U[a,b]\) (derivation also required)Yes
\(F(x)=\frac{x-a}{b-a}\), \(a\le x\le b\) (0 below, 1 above)cdf of \(U[a,b]\); quartiles and percentilesNo – derive it
\(E(g(X))=\int g(x)f(x)\,dx\); \(E(X^2)=\text{Var}(X)+[E(X)]^2\)expected value of a function of \(X\)Yes
rounding to nearest \(k\): error \(\sim U\left[-\frac{k}{2},\frac{k}{2}\right]\)modelling rounding errorsNo – derive it
\(B(n,p)\approx N(np,np(1-p))\)\(n\) large, \(p\) close to \(\frac12\)No – learn it
\(Po(\lambda)\approx N(\lambda,\lambda)\)\(\lambda\) largeNo – learn it
\(P(X\le k)\approx P(Y\lt k+0.5)\), \(P(X\ge k)\approx P(Y\gt k-0.5)\)continuity correctionNo – learn it
\(z=\frac{y-\mu}{\sigma}\)standardising before using \(\Phi(z)\)No – learn it

Check your understanding

1. \(X\sim U[-2,6]\). Find \(P(X\lt0)\) and \(\text{Var}(X)\).

\(P(X\lt0)=\frac28=0.25\). \(\text{Var}(X)=\frac{8^2}{12}=\frac{16}{3}\).

2. \(X\sim B(80,0.45)\). Write \(P(X\lt30)\) in terms of the approximating Normal variable \(Y\).

"Fewer than 30" means 29 or fewer, so \(P(Y\lt29.5)\), with \(Y\sim N(36,19.8)\).

3. Which approximation would you use for \(B(500,0.004)\), and why?

\(Po(2)\): \(n\) is large and \(p\) is small, and \(np=2\).

4. Why is a Normal approximation unsuitable for \(B(30,0.05)\)?

\(np=1.5\) 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. \(X\sim Po(64)\). Use a Normal approximation to find \(P(X\ge70)\).

\(Y\sim N(64,64)\); \(P(Y\gt69.5)=P(Z\gt5.5/8)=P(Z\gt0.69)=1-0.7549=0.2451\).

6. A length is recorded to the nearest 5 mm. Find the standard deviation of the rounding error.

The error is \(U[-2.5,2.5]\), width 5, so \(\sigma=\frac{5}{\sqrt{12}}=1.44\) mm (3 s.f.).

Specification checklist

  • (S2 3.1) I can write down the pdf of a continuous uniform distribution on \([a,b]\), with its range, and find probabilities from it.
  • (S2 3.1) I can derive and specify fully the cumulative distribution function of a uniform distribution and use it to find quartiles and percentiles.
  • (S2 3.1) I can derive the mean \(\frac{a+b}{2}\) and variance \(\frac{(b-a)^2}{12}\) by integration, and use them to find unknown parameters.
  • (S2 3.1) I can find \(E(g(X))\) for a uniform variable, including \(E(X^2)\) and linear functions.
  • (S2 3.1) I can model rounding errors and other "equally likely anywhere in a range" situations with a uniform distribution.
  • (S2 3.2) I can approximate \(B(n,p)\) by \(N(np,np(1-p))\) and state when this is suitable.
  • (S2 3.2) I can approximate \(Po(\lambda)\) by \(N(\lambda,\lambda)\) for large \(\lambda\), scaling the rate first where needed.
  • (S2 3.2) I can apply the correct continuity correction to any inequality or single value.
  • (S2 3.2) I can choose the most appropriate approximation (Poisson or Normal) for a given situation and justify the choice, and use an approximation in reverse with a percentage point to find an unknown.