Oct 2021: Paper 3 Section A: Statistics

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (a) many scored the mark by referring to the possibility of bias being introduced by the interviewer or the lack of randomness when using quota sampling. The most common answers that did not score the mark referred to cost or time failing to appreciate that quota sampling is generally quicker and therefore cheaper to implement than simple random sampling. Some thought that quota sampling required a sampling frame which suggested they were confused about the different types of sampling. Part (b) was generally answered

Common mistakes

• e.g. not random, (more likely to be) biased, cannot be used reliably for inferences
• time or cost (quota is quicker and cheaper), "not representative", "less accurate" or non-response score B0
• quota sampling needs no sampling frame

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

which suggested they were confused about the different types of sampling. Part (b) was generally answered very well with most stating the distribution, and scoring the method mark, and then evaluating part (i) correctly on their calculator. Part (ii) met with less success as many could not deal with the inequality or were unsure how to use their calculator correctly. Some seemed to confuse significant figures with decimal places: the front of the exam paper instructs then to use 3 significant figures but some only gave an answer of 0.022 here and lost this mark. Part (c) was usually answered correctly and many went on to use the

Common mistakes

• state B(36, 0.08)
• P(X ≥ 7) = 1 − P(X ≤ 6) = 0.0222
• give 3 s.f. (0.022 loses the mark)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

of 0.022 here and lost this mark. Part (c) was usually answered correctly and many went on to use the correct binomial distribution in part (d) though some found interpreting “fewer than 3” a problem and a

Common mistakes

• 40% of club members, so P(club and tango) = 0.4 × 0.08 = 0.032

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

correct binomial distribution in part (d) though some found interpreting “fewer than 3” a problem and a ). Overall though this proved a good start to the paper common incorrect answer was 0.924 (from P( X 3 and over 40% scored 6 or 7 marks here.

Common mistakes

• new model B(50, 0.032) using (c)
• fewer than 3 means P(T ≤ 2) = 0.785, not P(T ≤ 3) = 0.924

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most scored the mark in part (a) by identifying that the correlation was negative. When we ask students to “describe” we are simply requiring a non-contextual description. We will ask them to “interpret” when we require a contextual description. In this case we allowed students to describe the correlation in terms of the variables so answers such as “as x increases y decreases” were accepted. Some thought the correlation was strong and others weak but at this stage all we wanted was the negative description. Part (b) caused some

Common mistakes

• "negative" (strength not needed)
• "negative skew" is B0
• describe needs no context

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

strong and others weak but at this stage all we wanted was the negative description. Part (b) caused some problems with some stating that the scatter diagram showed negative correlation but failing to say whether or not this was compatible with Marc’s suggestion. Some students felt that there was no correlation shown by the scatter diagram and did not reconsider this even after correctly answering part (d).

Common mistakes

• must say the data are compatible with/support the suggestion AND give the reason (negative correlation)
• a reason based on a single point scores B0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most answered part (c) correctly though a number “lost” the minus sign between their calculator and writing down the answer on the page. Usually there were sufficient figures given for us to award the mark when the minus sign was included. In part (d) most attempted the hypotheses and the majority used ρ. Some

Common mistakes

• calculator r = −0.545
• do not lose the minus sign

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

by the scatter diagram and did not reconsider this even after correctly answering part (d). Most answered part (c) correctly though a number “lost” the minus sign between their calculator and when the minus sign was included. In part (d) most attempted the hypotheses and the majority used ρ. Some didn’t have “= 0” for their null hypothesis with alternative values or inequalities being used and of course some used r or simply wrote the hypotheses in words. We allowed (+) for the critical value provided the actual value was compatible with their alternative hypothesis and many achieved this mark but few scored the final mark as the conclusion was not related to the context of the question.

Common mistakes

• H₀: ρ = 0, H₁: ρ < 0 in terms of ρ (not r or words)
• 1-tail 5% critical value −0.4259
• conclusion in context (correlation between numbers of letters in last and first names)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although this question involved familiarisation with the large data set nearly 70% scored 3 or more marks here. Many students knew the units used for Daily Mean Pressure (or wrote “Pascals” which was allowed).

Common mistakes

• Daily Mean Pressure is in hectopascals (hPa)
• kPa or Pa alone not accepted

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although this question involved familiarisation with the large data set nearly 70% scored 3 or more marks here. Many students knew the units used for Daily Mean Pressure (or wrote “Pascals” which was allowed). The coding caused problems for many in part (b) finding y was easy enough but then adding the 1010 seemed to elude a great number of the candidates. Another frequently seen error was to add 1010 to Σy leading to the common wrong answer of 40.8. Part (c) was the most successful part with many correct

Common mistakes

• coding: x̄ = ȳ + 1010 = 214/30 + 1010 = 1017
• adding 1010 to Σy gives the common wrong 40.8

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although this question involved familiarisation with the large data set nearly 70% scored 3 or more marks here. Many students knew the units used for Daily Mean Pressure (or wrote “Pascals” which was allowed). leading to the common wrong answer of 40.8. Part (c) was the most successful part with many correct answers seen; sometimes the effect of coding was clearly stated but often, I suspect, it was simply ignored and the correct answer obtained anyway. Occasionally students felt they should add the 1010 on after reaching the correct value of 12.1 and this just lost the first method mark for this part. The final part proved

Common mistakes

• σx = σy as coding by subtraction does not change the s.d.
• √(5912/30 − 7.133²) = 12.1
• do not add 1010 afterwards

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Although this question involved familiarisation with the large data set nearly 70% scored 3 or more marks here. Many students knew the units used for Daily Mean Pressure (or wrote “Pascals” which was allowed). reaching the correct value of 12.1 and this just lost the first method mark for this part. The final part proved challenging. There was evidence that students knew the relative locations of the 3 sites and often they deduced that it was high pressure and therefore the winds circulated clockwise. The major problem though was that students did not seem to appreciate that Cardinal Wind Direction in the large data set gives the direction from which the wind comes. This led to an incorrect allocation of directions in the table, some students were clearly confused by this and felt that the circulation must be anticlockwise and then a correct allocation of wind directions could, of course, be deduced but this lost the first mark. A small percentage of students though did achieve full marks for this question.

Common mistakes

• 1028–1029 is about mean + s.d., so high pressure and clockwise
• locations north to south Leuchars, Heathrow, Hurn
• Cardinal Wind Direction is the direction the wind blows FROM: Heathrow NE, Hurn E, Leuchars W

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% scoring full marks. Most students navigated parts (a) and (b) successfully but a few missed the information that “No students read all three magazines” and wasted a lot of time trying to find two equations in p and q. The conditional probability in part (c) proved to be a challenge for some; there were those who effectively

Common mistakes

• exactly one magazine = 0.08 + 0.09 + 0.36 = 0.53

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% scoring full marks. Most students navigated parts (a) and (b) successfully but a few missed the information that “No students read all three magazines” and wasted a lot of time trying to find two equations in p and q. The conditional probability in part (c) proved to be a challenge for some; there were those who effectively

Common mistakes

• no student reads all three so p = 0
• P(G) = 0.25 gives q = 0.12
• missing the "no students read all three" information

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% The conditional probability in part (c) proved to be a challenge for some; there were those who effectively used P(E | S) and others were miscopying values from their Venn diagram. Those who failed to make any progress in (i) were often able to pick up a mark for a correct follow through when finding the value of t.

Common mistakes

• P(S given E) = (r + p)/(r + p + 0.09 + 0.05) = 5/12 gives r = 0.10
• not P(E given S)
• total probability 1 gives t = 0.20

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% Part (d) was the least successful part of the question often due to poor use of the notation. There were a curiously large number of students who confused events with the probability of events; for example we saw E′ E′ a number of cases of students writing S = 0.48 instead of P( S ) = 0.48. Another source of errors ∩ ∩ and was confusing the symbols ∪ in their expressions. There was a demand to show clear working here ∩ and this meant that we needed to see probabilities correctly labelled and an appropriate test clearly shown and a number of attempts lacked these features but there were plenty of good solutions to this part too.

Common mistakes

• P(S ∩ E′) = 0.48, P((S ∩ E′) ∩ G) = q = 0.12 = 0.48 × 0.25 so independent
• label probabilities P(…), not events
• confusing ∩ and ∪

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Most students used their calculators in part (a) with reasonable success but a frequent error was to use 0.1 instead of 0.01 and without any working being shown there was no opportunity to award the method mark.

Common mistakes

• standardise and use z = −2.3263 for 1% (not 0.1)
• k = 152.3
• show working (calculator-only wrong answers get no method mark)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (b) was a straightforward probability calculation designed to be completed on a calculator as the specification requires and so it was disappointing that nearly 25% of the candidates scored no marks at all on this question. Identifying the conditional probability in part (c) defeated many students and some of those

Common mistakes

• calculator normal probability P(150 < F < 175) = 0.915

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

this question. Identifying the conditional probability in part (c) defeated many students and some of those who seemed to be attempting to use a conditional probability were unable to use the correct notation to write their expression. Those who did start with a correct conditional probability and moved onto a ratio of probabilities often had a correct denominator but a common error was to have P(F > 160) as the numerator rather than P(160 < F < 175). Part (d) was a familiar hypothesis test and many students scored something

Common mistakes

• conditional: P(F > 160 given 150 < F < 175) = P(160 < F < 175)/P(150 < F < 175)
• numerator is not P(F > 160)
• 0.847

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

rather than P(160 < F < 175). Part (d) was a familiar hypothesis test and many students scored something here. The hypotheses were usually stated in terms of mu and most of the time 166.5 rather than 164.6 was used. Selecting the correct model caused a number of problems: some used 164.6 as the mean and others used 7.4 (or occasionally 6.1) as the standard deviation, simply stating the correct model will score the method mark here but a number of students didn’t show us the model they were using and we had to award this mark by implication from correct working. Most students successfully used the probability approach (as we would recommend) with a few attempting a critical region or using a z value and comparing with 1.6449 from the tables. The final mark required a correct interpretation of their calculation in context and some students simply stopped after stating that the result was significant and failed to score this final mark.

Common mistakes

• H₀: μ = 166.5, H₁: μ < 166.5
• X̄ ~ N(166.5, 7.4²/50), not mean 164.6 or s.d. 7.4 or 6.1
• P(X̄ < 164.6) = 0.0347 < 0.05
• conclusion in context about Mia's belief

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

There is a requirement in the specification for questions occasionally to draw together knowledge from different areas of 9MA0 and this was a question aiming to do that. Many students realised that to start the problem they could use the fact that the sum of the probabilities must equal 1 but fewer than half of the students could move beyond that mark. Those who realised they had a sum of logarithms were often able to arrive at the equation abc = 36 but few could devise a strategy to deduce the values of a, b and c. Some stated that all of the values had to be greater than 1 though it was not always clear that they deduced this from the given bullet points in the question. Around a quarter of the

Common mistakes

• sum of probabilities = 1 gives log₃₆(abc) = 1 so abc = 36
• each probability > 0 so a, b, c > 1
• distinct, so 2, 3, 6
• working must be shown (answer only 2/5)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

students found the correct set of values with varying degrees of explanation on the way. Part (b) was often left blank though the first mark was available without values for a, b and c provided brackets were used correctly for the expression. Nearly 10% managed to score full marks for the question.

Common mistakes

• P(X₁ = X₂) = sum of squared probabilities (log₃₆a)² + (log₃₆b)² + (log₃₆c)² = 0.381
• brackets needed