The normal distribution - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 20 exam-style questions on The normal distribution for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The scores, S, on a mock examination taken by a large number of students are modelled by \(S \sim N(52, 11^2)\). Students who score more than d marks are awarded a distinction. The teachers want 8% of the students to be awarded a distinction.

Using the model, find the value of d. Give your answer to 3 significant figures.

Model answer

\(P(S > d) = 0.08\), so \(P(S < d) = 0.92\)
\(\dfrac{d - 52}{11} = 1.4051\)
\(d = 52 + 11 \times 1.4051 = 67.456...\)
\(d = 67.5\) (3 s.f.)

Explanation

The top 8% of students score more than d, so d is the value with 92% of the distribution below it, which the inverse normal gives directly. Setting \(\dfrac{d - 52}{11}\) equal to the z-value 1.4051 earns the method mark. A common error is to use the lower tail, giving 36.5, or to use 0.08 as if it were a z-value.

Mark scheme

1 mark
M1 Standardises with d, 52 and 11 and sets equal to a z-value: \(\dfrac{d - 52}{11} = z\) with z in the range \(1.40 < |z| < 1.41\); or uses the inverse normal with \(P(S < d) = 0.92\); or a correct final answer of 67.5 with no working
An answer of 67.5 with no working scores 2 marks. The z-value is 1.4051 (from \(P(Z < z) = 0.92\)). Using \(z = -1.4051\) gives 36.5, which scores M1 A0. Using 0.08 or 0.92 as the z-value scores M0.
1 mark
A1 \(d = 67.5\) (awrt 67.5, i.e. [67.45, 67.55))
Exact value 67.4558... Isw a subsequent comment that a student needs 68 marks. A final answer of 68 alone (no 67.5 seen) scores A0, as the question asks for 3 significant figures.

Question 2

4 marks

A supermarket delivers groceries in one-hour time slots. For a delivery booked in a slot, the time, A minutes, after the start of the slot at which the van arrives is modelled by \(A \sim N(32, 14^2)\). A negative value of A means that the van arrives before the slot starts. A delivery is late if \(A > 60\).

Given that a particular delivery is not late, find the probability that it arrives during the first 15 minutes of its slot. Give your answer to 3 significant figures.

Model answer

\(P(0 < A < 15 \mid A < 60) = \dfrac{P(0 < A < 15)}{P(A < 60)}\)
\(P(0 < A < 15) = 0.10118...\)
\(P(A < 60) = 0.97725...\)
Probability \(= \dfrac{0.10118...}{0.97725...} = 0.10354... = 0.104\) (3 s.f.)

Explanation

"Given that it is not late" restricts attention to \(A < 60\), so divide the probability of arriving in the first 15 minutes, \(P(0 < A < 15)\), by \(P(A < 60)\); the event in the numerator already lies inside \(A < 60\). Both probabilities come from a calculator. Common errors are to use \(P(A < 15)\), which wrongly includes arrivals before the slot, and to forget to divide.

Mark scheme

1 mark
M1 Correct conditional probability ratio: \(\dfrac{P(0 < A < 15)}{P(A < 60)}\); or a correct final answer of 0.104 with no working
An answer of 0.104 with no working scores 4 marks (accept awrt 0.104, i.e. [0.1035, 0.1045)). The numerator must be \(P(0 < A < 15)\) and the denominator \(P(A < 60)\). Using \(P(A < 15)\) in the numerator (giving 0.115) scores M0 A0 but can score the B1 for the denominator.
1 mark
B1 \(P(0 < A < 15) = 0.101\) (awrt 0.101, i.e. [0.1005, 0.1015))
Exact value 0.10118...
1 mark
B1 \(P(A < 60) = 0.977\) (awrt 0.977, i.e. [0.9765, 0.9775))
Exact value 0.97725...
1 mark
A1 0.104 (awrt 0.104, i.e. [0.1035, 0.1045))
Depends on the M1. Exact value 0.10354... Allow 0.103 if it comes from correct probabilities rounded to 3 s.f. (0.101 ÷ 0.977).

Question 3

6 marks

An airline weighs the cabin bags of passengers at the gate. The mass, W kg, of a passenger's cabin bag is modelled by \(W \sim N(9.1, 1.4^2)\). At present, a passenger whose bag weighs more than 10 kg pays a fee of £15.

The airline plans a new policy in which only a passenger whose bag weighs more than t kg pays a fee, of £40. Find the value of t for which the expected fee paid per passenger is the same under both policies. Give your answer to 3 significant figures.

Model answer

Present policy: \(P(W > 10) = P\left(Z > \dfrac{10 - 9.1}{1.4}\right) = P(Z > 0.6429) = 0.26015...\)
Expected fee per passenger \(= 15 \times 0.26015... = £3.902\)
New policy: \(40 \times P(W > t) = 3.902\), so \(P(W > t) = 0.097559...\)
\(\dfrac{t - 9.1}{1.4} = 1.2956\)
\(t = 9.1 + 1.4 \times 1.2956 = 10.913...\)
t = 10.9 kg (3 s.f.)

Explanation

The expected fee per passenger is the fee multiplied by the probability of paying it, so the present policy gives \(15 \times 0.260 = £3.90\). Setting \(40 \times P(W > t)\) equal to this gives \(P(W > t) = 0.0976\), and the inverse normal turns that into t = 10.9 kg. A common error is to use the probability 0.0976 itself as a z-value.

Mark scheme

1 mark
M1 Uses \(N(9.1, 1.4^2)\) to find \(P(W > 10)\); or a correct final answer of t = 10.9 with no working
An answer of t = 10.9 with no working scores 1 mark (the final A1). Accept awrt 10.9.
1 mark
A1 \(P(W > 10) = 0.260\) (awrt 0.260, i.e. [0.2595, 0.2605))
Exact value 0.26015... The present expected fee per passenger is £3.90 (awrt 3.90, i.e. [3.895, 3.905)).
1 mark
M1 Forms \(40 \times P(W > t) = 15 \times P(W > 10)\), with their \(P(W > 10)\)
Accept the expected fees compared for any number of passengers.
1 mark
A1 \(P(W > t) = 0.0976\) (awrt 0.0976, i.e. [0.09755, 0.09765))
Exact value 0.097559... (\(= 0.375 \times 0.26015\)).
1 mark
M1 Uses the inverse normal: \(\dfrac{t - 9.1}{1.4} = z\) with z in the range \(1.29 < z < 1.30\), or \(P(W < t) = 0.9024\) on a calculator
Follow through their \(P(W > t)\). Using 0.0976 or 0.9024 as a z-value scores M0.
1 mark
A1 \(t = 10.9\) (awrt 10.9, i.e. [10.85, 10.95))
Exact value 10.913...

Question 4

9 marks

A machine makes steel washers. The diameter, D mm, of a washer is modelled by \(D \sim N(\mu, 0.05^2)\), where \(\mu\) can be set by an engineer. A washer is rejected if its diameter is less than 9.9 mm (too small) or more than 10.1 mm (too large).

The machine is set with \(\mu = 10.03\). Find the proportion of washers that are rejected. A washer is chosen at random from those that are rejected. Find the probability that it is too large.

The engineer changes only the value of \(\mu\), so that 1% of the washers are too large. Find the new value of \(\mu\), giving your answer to 3 decimal places, and find the new proportion of washers that are rejected, giving your answer to 2 significant figures.

Give your other answers to 3 significant figures.

Model answer

\(\mu = 10.03\): \(P(D < 9.9) = P(Z < -2.6) = 0.00466\) and \(P(D > 10.1) = P(Z > 1.4) = 0.08076\)
Proportion rejected \(= 0.00466 + 0.08076 = 0.0854\) (3 s.f.)
\(P(\text{too large} \mid \text{rejected}) = \dfrac{0.08076}{0.08542} = 0.945\) (3 s.f.)
New mean: \(P(D > 10.1) = 0.01\), so \(\dfrac{10.1 - \mu}{0.05} = 2.3263\) and \(\mu = 10.1 - 2.3263 \times 0.05 = 9.98368...\) = 9.984 (3 d.p.)
\(P(D < 9.9) = P\left(Z < \dfrac{9.9 - 9.98368}{0.05}\right) = P(Z < -1.6737) = 0.04710\)
New proportion rejected \(= 0.04710 + 0.01 = 0.057\) (2 s.f.)

Explanation

With \(\mu = 10.03\) most rejected washers are too large, so given that a washer is rejected it is very likely (0.945) to be too large: divide \(P(D > 10.1)\) by the total proportion rejected. Setting 1% too large puts 10.1 at 2.3263 standard deviations above the new mean, 9.984 mm; this raises the proportion too small to 4.7% but the total rejected falls from 8.5% to 5.7%. Common errors are using \(P(D > 10.1)\) alone for the conditional probability and rounding \(\mu\) to 9.98 before finding the new proportion.

Mark scheme

1 mark
M1 Uses \(N(10.03, 0.05^2)\) to find \(P(D < 9.9) + P(D > 10.1)\) (or \(1 - P(9.9 < D < 10.1)\)); correct final answers of 0.0854, 0.945, 9.984 and 0.057 (or 0.056) with no working score 4 marks
Answers of 0.0854, 0.945, 9.984 and 0.057 (or 0.056) with no working score 4 marks (the four A1 marks): the methods for the conditional probability and the new mean must be seen for the other marks. Tails: \(P(D < 9.9) = 0.00466\) and \(P(D > 10.1) = 0.0808\).
1 mark
A1 Proportion rejected = 0.0854 (awrt 0.0854, i.e. [0.08535, 0.08545))
Exact value 0.085417... Allow 8.54%.
1 mark
M1 \(P(\text{too large} \mid \text{rejected}) = \dfrac{P(D > 10.1)}{P(\text{rejected})}\) with their values
Accept their values, unrounded or to 3 s.f. The numerator must be \(P(D > 10.1)\) and the denominator their proportion rejected. Using \(P(D > 10.1)\) alone (0.0808) scores M0.
1 mark
A1 0.945 (awrt 0.945 or 0.946, i.e. [0.9445, 0.9465))
Exact value 0.94543...; 0.946 from \(0.0808 \div 0.0854\) is accepted.
1 mark
B1 Uses \(z = 2.3263\) (from \(P(Z > z) = 0.01\))
Accept 2.326 or better.
1 mark
M1 \(\dfrac{10.1 - \mu}{0.05} = z\) with z in the range \(2.3 < |z| < 2.4\)
Independent of the B1. Using 0.01 or 0.99 as a z-value scores M0.
1 mark
A1 \(\mu = 9.984\) (awrt 9.984, i.e. [9.9835, 9.9845))
Exact value 9.98368... Needs compatible signs (\(\mu\) is below 10.1).
1 mark
M1 Finds \(P(D < 9.9)\) using \(N(\text{their } \mu, 0.05^2)\) and adds 0.01 (or adds their \(P(D > 10.1)\))
Accept a correct method with their new \(\mu\) (provided it is less than 10.03). \(P(D < 9.9) = 0.0471\) from the unrounded \(\mu\), 0.0465 from \(\mu = 9.984\).
1 mark
A1 New proportion rejected = 0.057 (awrt 0.057, i.e. [0.0565, 0.0575)), or 0.056 from \(\mu = 9.984\)
Exact value 0.057099... from the unrounded \(\mu\). Using the 3 d.p. value \(\mu = 9.984\) gives 0.05648..., which is 0.056 to 2 s.f.: accept 0.056 (or 0.0565) from this route. Using \(\mu = 9.98\) (0.0648, i.e. 0.065) scores A0.

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