Oct 2021: AS AS Paper 2 Section A: Statistics

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

Examiner's comment

The vast majority were able to find the correct value for p but under half managed to identify a pair of mutually exclusive events. Some lost the mark for confusing events with probabilities of events and others thought the answer was A and B or B and C.

Common mistakes

• probabilities in a Venn diagram sum to 1
• p = 1 − 0.7 = 0.3

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

Examiner's comment

The vast majority were able to find the correct value for p but under half managed to identify a pair of mutually exclusive events. Some lost the mark for confusing events with probabilities of events and others thought the answer was A and B or B and C.

Common mistakes

• answer is the events A and C, not P(A) and P(C) or A ∩ C = ∅
• A and B or B and C (overlapping) are common wrong answers
• C inside B is not exclusive

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

Examiner's comment

The response to this question on histograms was much better than in previous series with over 50% scoring 7 or more marks. Most managed to complete at least one of the bars of the histogram correctly but a

Common mistakes

• find the scale from the [5, 20) bar: fd = 45/15 = 3
• bars for [0, 5) at fd 1 and [20, 40) at fd 4.5
• area represents frequency

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

Examiner's comment

The response to this question on histograms was much better than in previous series with over 50% scoring 7 or more marks. Most managed to complete at least one of the bars of the histogram correctly but a sizeable minority did not make any progress at all here. In part (b) many were able to complete the table correctly and find the total number of passengers. They could often go on to find the median but there were some errors in class boundaries and a fair number of very confused attempts. In part (c) most were able to

Common mistakes

• read fd from the histogram to get 130 and 60, total 331
• median from the 40–65 class: 40 + (165.5 − 140)/130 × 25 = 44.9
• errors in class boundaries

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

Examiner's comment

The response to this question on histograms was much better than in previous series with over 50% scoring 7 or more marks. Most managed to complete at least one of the bars of the histogram correctly but a some errors in class boundaries and a fair number of very confused attempts. In part (c) most were able to find the upper outlier limit but some were not able to explain clearly why they could then deduce that none of the passengers were outliers. We required a clear statement that none of the passengers were over 90 and 90 was less than this limit so there were no outliers.

Common mistakes

• upper limit 58.9 + 1.5 × 31.6 = 106.3
• must state that no passenger is 90 or over and 90 < 106 so the oldest is not an outlier

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

Examiner's comment

Although some showed a genuine familiarity with the large data set many seemed totally unfamiliar large data set, most correctly identified the sampling technique in part (a) and some drew a sensible inference in part (c) by simply stating that variable A was the most frequent.

Common mistakes

• systematic sampling (every 10th reading)
• "systemic" not accepted

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

Examiner's comment

Although some showed a genuine familiarity with the large data set many seemed totally unfamiliar with it and subsequently could not make progress in part (b). For part (b) the students needed to be familiar with the qualitative variables in the large data set (wind speed and cardinal direction were the only choices) and then select an appropriate variable in the knowledge that there were only 3 outcomes. Those who were familiar with the data set would then know that the majority of wind speeds are “light”. Aside from the large data set, most correctly identified the sampling technique in part (a) and some drew a sensible

Common mistakes

• qualitative variables in the LDS are wind speed (Beaufort descriptions) and cardinal wind direction
• only 3 outcomes means daily mean wind speed
• modal outcome A is "light"
• "wind" or "wind direction" not accepted

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

Examiner's comment

Although some showed a genuine familiarity with the large data set many seemed totally unfamiliar inference in part (c) by simply stating that variable A was the most frequent.

Common mistakes

• the samples estimate the frequency of A reliably (A occurs most, about 80–90% of the time)
• do not try to infer B and C from such small counts

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

Examiner's comment

Most students were able to identify the correct binomial model in part (a) and use this to find the probabilities in part (b). However nearly 15% of students failed to score any marks in this question and this

Common mistakes

• B(18, 0.14)

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

Examiner's comment

probabilities in part (b). However nearly 15% of students failed to score any marks in this question and this is possibly a reflection of the weaker nature of the entry this time. In part (c) students were expected to

Common mistakes

• (i) P(R = 1) = 0.194
• (ii) at least 4 means 1 − P(R ≤ 3) = 0.238, not 1 − P(R ≤ 4)

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

Examiner's comment

is possibly a reflection of the weaker nature of the entry this time. In part (c) students were expected to comment on the suitability of this binomial model and we were looking for a recognition that removing beads without replacement would impact on the assumption that p is constant and so a large number of beads in the sack would be needed to ensure that the binomial distribution could still be used as a model; sadly none of the students scored the mark here.

Common mistakes

• p = 0.14 must stay constant: removing beads without replacement changes it unless the sack holds a large number of beads
• "events are independent" alone not accepted
• almost no candidate scored this

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

Examiner's comment

Part (d) was a standard binomial hypothesis test and most students stated correct hypotheses and a correct model. There was still some confusion about the appropriate probability to calculate and a significant number found P(X = 4) rather than P( X „ 4). Those who did find the correct probability were usually able to give a correct conclusion in context which is encouraging however few knew what the p- value was in part (e).

Common mistakes

• two-tailed: H₀ p = 0.14, H₁ p ≠ 0.14
• X ~ B(75, 0.14)
• P(X ≤ 4) = 0.0151 compared with 0.025, not P(X = 4)
• conclusion in context: evidence the proportion of red beads has changed

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

Examiner's comment

value was in part (e).

Common mistakes

• two-tailed test so p-value = 2 × 0.0151 = 0.03
• few knew what the p-value was

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

Examiner's comment

Sadly two thirds of the candidates were not able to make any progress with this question. Those who did pause for a moment and consider the situation often realised that there were two possible cases to consider. The majority of these failed to give the correct probabilities though as they did not realise that after the initial transfer from bag A there were 10, rather than 9, balls in bag B. A handful of the students did negotiate the problem successfully though some then added together their probabilities (instead of giving the two pairings of n and p) and lost the final mark.

Common mistakes

• two cases: 3 of each colour (n = 2, red then green, p = 2/45) or 4 of each (n = 5, green then yellow, p = 1/8)
• bag B has 10 balls after the first transfer, not 9
• give the pairs, do not add the probabilities