The normal distribution

• Sketch first, every time. Divide by \(\sigma\), never by \(\sigma^2\). A right tail is \(1 - \Phi\). A left tail smaller than one half sits at a negative \(z\).
• \(Z = \dfrac{X - \mu}{\sigma}\) with the variance second in \(\mathrm{N}(\mu, \sigma^2)\); a right tail is \(1 - \Phi\) and a left tail below one half has a negative \(z\); \(np > 5\) and \(nq > 5\), then move every boundary half a unit outwards.

• 5.5.1 I understand the use of a normal distribution to model a continuous random variable, and use normal distribution tables
• 5.5.2 I can solve problems concerning a variable X, where X ~ N(μ, σ²), including (a) finding the value of P(X > x₁), or a related probability, given the values of x₁, μ, σ (b) finding a relationship between x₁, μ and σ given the value of P(X > x₁) or a related probability
• 5.5.3 I can recall conditions under which the normal distribution can be used as an approximation to the binomial distribution, and use this approximation, with a continuity correction, in solving problems

Accuracy for this chapter. \(z\) to three decimal places (so the ADD column can be used), \(\Phi\) to four decimal places exactly as printed, the final probability to 3 significant figures. Keep \(\sigma\) unrounded when it comes from a square root or a division (\(\sqrt{16.8} = 4.0988\), \(\dfrac{15}{2.927} = 5.1247\)) and round once at the end. Quote critical values as printed: 1.960, not 1.96. The table stops at 2.99; beyond it, say \(\Phi(z) > 0.9986\) rather than inventing a value. The syllabus asks for full details of the standardisation and states that no marks are given for unsupported answers from a calculator, so the line with the numbers in it is always written.

• “\(X \sim \mathrm{N}(50, 16)\), so \(\sigma = 16\).” Correct The second parameter is the variance. \(\sigma^2 = 16\), so \(\sigma = 4\), and the standardisation is \(z = \dfrac{x - 50}{4}\). Read the brackets as \(\mathrm{N}(\text{mean}, \text{variance})\) every time; \(\mathrm{N}(800, 12^2)\) is the same convention with the square left visible.
• “\(\mathrm{P}(X > 55) = \Phi(1.25) = 0.8944\).” Correct \(\Phi\) is always the area to the left. A right tail is \(1 - \Phi(z) = 1 - 0.8944 = 0.1056\). The sketch catches it: a thin tail cannot have area 0.89.
• “\(\Phi(-1.25) = -0.8944\).” Correct The table has no negative \(z\). By symmetry \(\Phi(-z) = 1 - \Phi(z)\), so \(\Phi(-1.25) = 0.1056\). A probability is never negative.
• “\(\Phi(1.233) = 0.8907\).” Correct 0.8907 is \(\Phi(1.23)\). The third decimal comes from the ADD column: row 1.2, ADD column 3 is 6, so \(\Phi(1.233) = 0.8907 + 0.0006 = 0.8913\).
• “\(\mathrm{P}(X > 70) = 0.2\), so \(\dfrac{70 - \mu}{5} = 0.2\).” Correct 0.2 is an area, not a \(z\)-value. Turn the area into \(z\) first: the area to the left of 70 is 0.8, and \(\Phi(z) = 0.8\) gives \(z = 0.842\).
• “\(\mathrm{P}(X < 20) = 0.1\), so \(\dfrac{20 - \mu}{\sigma} = 1.282\).” Correct A left tail smaller than 0.5 lies below the mean, so its \(z\) is negative: \(-1.282\).
• “\(X \sim \mathrm{B}(100, 0.5)\), so \(\mathrm{P}(X \ge 60) \approx \mathrm{P}(Y > 60)\).” Correct Each integer is the interval half a unit either side of it, and 60 is included, so the boundary moves out to 59.5: \(\mathrm{P}(Y > 59.5)\). And state both conditions, \(np > 5\) and \(nq > 5\), before you approximate.
• “\(\mathrm{N}(50, 16)\), so \(\sigma = 16\).” Repair The second parameter is the variance: \(\sigma = \sqrt{16} = 4\).
• “\(z = \dfrac{55 - 50}{16}\).” Repair Divide by the standard deviation, not the variance: \(z = \dfrac{55 - 50}{4} = 1.25\).
• “\(\mathrm{P}(X < 60)\) is less than \(\mathrm{P}(X \le 60)\).” Repair For a continuous variable \(\mathrm{P}(X = 60) = 0\), so the two are equal.
• “\(\mathrm{P}(X > 55) = \Phi(1.25) = 0.8944\).” Repair \(\Phi\) is a left area; a right tail is \(1 - \Phi(1.25) = 0.1056\). Sketch and shade before you decide whether to subtract.
• “\(\Phi(-1.25) = -0.8944\).” Or: “0.8907 for \(z = 1.233\).” Repair \(\Phi(-z) = 1 - \Phi(z)\), so \(\Phi(-1.25) = 0.1056\): a probability is never negative. And a third decimal place needs the ADD column: \(0.8907 + 0.0006 = 0.8913\).
• “\(z = 1.2333\), so I will use \(z = 1.23\).” Or rounding \(z\) to 1.3. Repair Keep three decimal places and use the ADD column. Rounding \(z\) moves the answer, often in the second or third significant figure.
• “\(\mathrm{P}(X > 70) = 0.2\), so \(\dfrac{70 - \mu}{5} = 0.2\).” Repair 0.2 is an area, not a \(z\)-value. The left area is 0.8, and \(\Phi(z) = 0.8\) gives \(z = 0.842\).
• Using \(z = 1.960\) for a one-tailed probability of 0.05. Repair A right tail of 0.05 is a left area of 0.95, and \(\Phi(z) = 0.95\) gives \(z = 1.645\). 1.960 belongs to \(\Phi(z) = 0.975\): a tail of 0.025, or a symmetric middle of 0.95.
• “\(\mathrm{P}(X < 20) = 0.1\), so \(\dfrac{20 - \mu}{\sigma} = 1.282\).” Repair A left tail smaller than 0.5 is below the mean, at a negative \(z\): \(-1.282\).
• “From \(\mathrm{P}(X < 30) = 0.9\) alone, \(\mu = 30 - 1.282 = 28.718\).” Repair The equation is \(30 - \mu = 1.282\sigma\): the \(z\)-value multiplies \(\sigma\). With \(\sigma\) unknown this is only a relationship; a second probability or the value of \(\sigma\) is needed.
• “\(\mathrm{B}(100, 0.5) \approx \mathrm{N}(50, 5)\).” Repair The variance is \(npq = 25\): write \(\mathrm{N}(50, 25)\), and use \(\sigma = 5\) in the standardisation.
• “\(\mathrm{P}(X \ge 60) \approx \mathrm{P}(Y > 60)\).” Repair The continuity correction makes it \(\mathrm{P}(Y > 59.5)\). Without it the answer is 0.0228 instead of 0.0287: wrong in the second significant figure.
• “\(\mathrm{P}(X > 60) \approx \mathrm{P}(Y > 59.5)\).” Repair \(X > 60\) means \(X \ge 61\), whose bar starts at 60.5: \(\mathrm{P}(Y > 60.5)\).
• Applying the normal approximation to \(\mathrm{B}(20, 0.15)\), or stating only “\(n\) is large”. Repair State and check both conditions with numbers: here \(np = 3\), not greater than 5, so the approximation is not valid and the binomial formula is used. “\(n\) is large” is not the condition.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. Chapter 28 approximates the Poisson by a normal with the same continuity correction: re-answer worked example 7 then. Chapter 29 combines normal variables, and every answer there ends in a standardisation from this chapter: re-answer retrieval question 7. Chapter 31 makes the sample mean normal, and chapter 32 uses the critical values of 5.5.2 as test boundaries: re-answer worked example 6 before each. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Chapter 27 · Probability & Statistics 1 · How it is assessed
• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Probability & Statistics 1 content, examined in Paper 5. Paper 5 (Probability & Statistics 1) is 40% of an AS Level that includes it and 20% of the A Level, for which it is compulsory. Its questions use no algebraic methods beyond the Paper 1 content, and it is the foundation for Paper 6. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A model stated in context, then a forward probability (sometimes with a sketch asked for first); an inverse part that gives a probability and asks for a boundary, the mean or the standard deviation, or both from two probabilities; a binomial situation where the conditions must be justified before the normal approximation is used with a continuity correction. Parts can be chained: a probability found in one part becomes the \(p\) of a binomial in the next.
• MF19 prints the \(\Phi\) table with its ADD columns, the critical-value table, and the binomial mean \(np\) and variance \(np(1 - p)\). Standardisation, the four shapes, the approximation conditions and the continuity correction are not printed and must be known (MF19 card).
• The syllabus asks for full details of the standardisation, so write \(z = \dfrac{x - \mu}{\sigma}\) with the numbers in it every time. \(z\) to three decimal places, \(\Phi\) to four, the answer to 3 significant figures; \(\sigma\) kept unrounded through the working. Both approximation conditions stated with their values.