Pearson Edexcel International Advanced Level Mathematics: The Normal distribution (YMA01) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Specification Issue 3, April 2019, section 6 of Statistics 1 (numbered 5.1 in the specification). The specification states that S1 is an optional unit for IAS and IAL Mathematics and Further Mathematics. Where a question needs a value from a table, it is given in the question. Give answers to 3 significant figures unless a question says otherwise.

Study first with the study guide and the revision notes. Course pages: the Edexcel A Level Mathematics hub and the printable checklist.

Questions

1. The random variable W ~ N(25, 4). Write down the standard deviation of W. Given that Φ(1) = 0.8413, find P(W < 27). [3]

2. The time, J minutes, taken by a bus to complete a route is modelled by J ~ N(42, 6²). Given that Φ(1) = 0.8413 and Φ(1.5) = 0.9332:

(a) find P(J > 48), [2] (b) find P(36 < J < 51). [3]

3. The length L mm of a steel rod is modelled by L ~ N(120, 0.5²). Given that Φ(z) = 0.95 gives z = 1.6449, find the value of k such that P(L < k) = 0.05. Give k to 2 decimal places. [3]

4. In a test, the mark M of a candidate is modelled by M ~ N(55, σ²). It is given that P(M > 70) = 0.0668. Given that Φ(1.5) = 0.9332 and Φ(1) = 0.8413:

(a) show that σ = 10, [3] (b) estimate how many of 1000 candidates score a mark below 45. [2]

5. A machine fills bottles. The volume V ml in a bottle is modelled by V ~ N(μ, 3²). The machine is set so that P(V < 500) = 0.01. Given that Φ(z) = 0.99 gives z = 2.3263, find μ. [4]

6. The random variable X ~ N(μ, σ²) satisfies P(X < 20) = 0.10 and P(X > 35) = 0.25. Given that Φ(z) = 0.90 gives z = 1.2816 and Φ(z) = 0.75 gives z = 0.6745, find μ and σ. [6]

7. The random variable X is Normally distributed with P(X < 12) = P(X > 20) = 0.1587. Given that Φ(1) = 0.8413, find the mean and standard deviation of X. [4]

8. (a) State two properties of the curve of a Normal distribution. [2] (b) Z ~ N(0, 1) and Φ(0.4) = 0.6554. Find P(Z < −0.4). [2]

9. X ~ N(100, 15²). Given that Φ(z) = 0.95 gives z = 1.6449, find the value of a such that P(100 − a < X < 100 + a) = 0.90. [4]

10. Bags of flour are packed by a machine. The mass W grams of a bag is modelled by W ~ N(μ, σ²). It is found that P(W < 495) = 0.0228 and P(W > 510) = 0.1587. Given that Φ(2) = 0.9772, Φ(1) = 0.8413 and Φ(0.6) = 0.7257:

(a) find μ and σ, [6] (b) find P(W < 500), [2] (c) find P(W > 508). [2]

11. A factory makes rods whose diameter D mm is modelled by D ~ N(10, 0.04²). A rod is rejected if D < 9.9 or D > 10.1. Use Φ(2.5) = 0.9938, Φ(1.75) = 0.9599, Φ(3.25) = 0.9994, and Φ(z) = 0.995 gives z = 2.5758.

(a) Find the probability that a rod is rejected. [3] (b) A fault moves the mean to 10.03 mm, with the standard deviation unchanged. Find the new probability that a rod is rejected. [4] (c) With the mean back at 10 mm, find the standard deviation needed so that the probability of rejection is 0.01. [3]

Answers

1. Standard deviation = √4 = 2 [1]. z = (27 − 25)/2 = 1 [1]. P(W < 27) = Φ(1) = 0.8413 [1] [3] Examiner insight: The second number in N(25, 4) is the variance. Writing the square root explicitly shows where the 2 comes from.

2. (a) z = (48 − 42)/6 = 1 [1]. P(J > 48) = 1 − 0.8413 = 0.1587 [1] (b) z values −1 and 1.5 [1]. P = Φ(1.5) − (1 − Φ(1)) = 0.9332 + 0.8413 − 1 [1] = 0.7745 [1] [5] Examiner insight: Show both standardised values on separate lines. A single slip with a sign is easier to spot when the standardising is visible.

3. P(L < k) = 0.05 is below 0.5, so z = −1.6449 [1]. (k − 120)/0.5 = −1.6449, so k = 120 − 1.6449 × 0.5 [1]. k = 119.17755, so k = 119.18 [1] [3] Examiner insight: A value for the lower 5% must sit below the mean of 120. If your answer is above it, the sign of z is wrong.

4. (a) P(M < 70) = 1 − 0.0668 = 0.9332 = Φ(1.5) [1]. (70 − 55)/σ = 1.5 [1]. σ = 15/1.5 = 10 [1] (b) z = (45 − 55)/10 = −1, so P(M < 45) = 1 − 0.8413 = 0.1587 [1]. Expected number = 1000 × 0.1587 = 158.7, so about 159 [1] [5] Examiner insight: On a “show that” question, state the step that links the given probability to Φ(1.5). Finishing with a bare 10 leaves the reasoning unseen.

5. P(V < 500) = 0.01 is below 0.5, so z = −2.3263 [1]. (500 − μ)/3 = −2.3263 [1]. μ = 500 + 2.3263 × 3 = 506.9789 [1]. μ = 507 (3 s.f.) [1] [4] Examiner insight: This question checks that you attach the minus sign. A set mean must be above 500 to make the lower tail small, so a value under 500 signals the sign error.

6. P(X < 20) = 0.10, so z = −1.2816 [1]. P(X < 35) = 0.75, so z = 0.6745 [1]. Equation 1: 20 − μ = −1.2816σ [1]. Equation 2: 35 − μ = 0.6745σ [1]. Subtract: 15 = 1.9561σ, so σ = 7.67 (3 s.f.) [1]. μ = 20 + 1.2816 × 7.6683… = 29.83, so μ = 29.8 (3 s.f.) [1] [6] Examiner insight: P(X > 35) = 0.25 must be converted to a left-tail probability of 0.75 before you use it. Keep the unrounded σ when you find μ.

7. The two values are equally far from the mean, so μ = (12 + 20)/2 = 16 [1], by symmetry [1]. P(X < 12) = 0.1587 = 1 − Φ(1), so the z-value is −1 [1]. (12 − 16)/σ = −1, so σ = 4 [1] [4] Examiner insight: Giving the reason for the mean, symmetry about x = μ, is better than a bare 16. Avoid solving two equations when the shortcut is available.

8. (a) Any two from: bell-shaped; symmetrical about the mean; mean, median and mode are equal; the total area under the curve is 1. [1] [1] (b) P(Z < −0.4) = 1 − Φ(0.4) [1] = 1 − 0.6554 = 0.3446 [1] [4] Examiner insight: “Two properties” needs two distinct points. “Bell-shaped” and “symmetrical” are separate. Repeating one property in different words earns no more.

9. The tails are equal, so each is 0.05 [1]. P(X < 100 + a) = 0.95 [1]. a/15 = 1.6449 [1]. a = 24.6735, so a = 24.7 (3 s.f.) [1] [4] Examiner insight: A central 90% interval leaves 5% in each tail, so you look up 0.95, not 0.90. This is the most common slip on this type.

10. (a) P(W < 495) = 0.0228 = 1 − Φ(2), so z = −2 [1]. P(W > 510) = 0.1587, so P(W < 510) = 0.8413 = Φ(1) and z = 1 [1]. Equation 1: 495 = μ − 2σ [1]. Equation 2: 510 = μ + σ [1]. Subtract: 15 = 3σ, so σ = 5 [1]. μ = 510 − 5 = 505 [1] (b) z = (500 − 505)/5 = −1 [1]. P(W < 500) = 1 − 0.8413 = 0.1587 [1] (c) z = (508 − 505)/5 = 0.6 [1]. P(W > 508) = 1 − 0.7257 = 0.2743 [1] [10] Examiner insight: Parts (b) and (c) follow on from (a). Quote your values for μ and σ at the start of each, so the working is easy to follow.

11. (a) z = (9.9 − 10)/0.04 = −2.5 and z = (10.1 − 10)/0.04 = 2.5 [1]. P(rejected) = 2 × (1 − 0.9938) [1] = 0.0124 [1] (b) z = (10.1 − 10.03)/0.04 = 1.75 and z = (9.9 − 10.03)/0.04 = −3.25 [1]. P(D > 10.1) = 1 − 0.9599 = 0.0401 [1]. P(D < 9.9) = 1 − 0.9994 = 0.0006 [1]. Total = 0.0407 [1] (c) Each tail has probability 0.005, so Φ(z) = 0.995 [1]. 0.1/σ = 2.5758 [1]. σ = 0.03882…, so σ = 0.0388 mm (3 s.f.) [1] [10] Examiner insight: In (b) the two tails are no longer equal, so you must find both and add them. Doubling one tail, as in (a), gives a wrong value once the mean moves.

Where marks are usually lost

• Using the variance as if it were the standard deviation when standardising.
• Choosing the wrong tail: writing Φ(z) where the question asks for P(Z > z).
• Dropping the minus sign on z in inverse questions where the probability is below 0.5.
• Rounding z or σ early in a simultaneous-equation question.
• Looking up 0.90 instead of 0.95 for a symmetrical 90% interval.
• Not writing the standardising step, which leaves a slip hard to trace.
• Giving a probability above 1 or an expected count with no rounding comment.
• Missing the symmetry shortcut when two probabilities are equal.

Next steps

• Re-read the study guide for any question that cost you marks.
• Use the revision notes for a quick recap.
• Open the Edexcel A Level Mathematics hub and tick the topic off on the checklist.
• Take the free 10-minute diagnostics to find weak spots.
• Book a free trial class if you would like a teacher to go through the questions with you.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics – Specification, Issue 3, April 2019, first teaching September 2018, published by Pearson Education Limited.