June 2023: AS AS Paper 2 Section A: Statistics

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

Examiner's comment

Part (a) was a generally successful start to the paper with the vast majority of students able to show sufficient calculations to arrive at the given mean. A few students opted to use the upper class boundary in their calculation. On some occasions 3186 50 was written with no supporting evidence. Some students misunderstood the demand and attempted to use interpolation to find the median instead of the mean.

Common mistakes

• read the histogram: frequency = frequency density × class width 2 (3, 12, 8, 2 doubled to 6, 24, 16, 4)
• use midpoints 61, 63, 65, 67, not upper boundaries
• 3186/50 alone with no products scores 0
• some found the median instead

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

Examiner's comment

On the whole part (b) was also well attempted. A common error remains that students confuse ∑𝑓𝑥2 with (∑𝑓𝑥)2.

Common mistakes

• sd = √(Σfx²/50 − 63.72²) with Σfx² = 203138 gives 1.59 (s = 1.60 allowed)
• confusing Σfx² with (Σfx)²

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

Examiner's comment

Many students were able to score the mark in part (c) as they gave an acceptable explanation why the standard deviation does not change. Some said it did not change but provided no supporting explanation whilst others believed the change in the mean would affect the standard deviation.

Common mistakes

• no effect: subtracting 5 g from every weight does not change the spread (only multiplying/dividing does)
• a reason is needed
• some thought the change in mean changes the sd

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

Examiner's comment

This question was the second most accessible question on the paper thanks to part (a) and (b), but due to parts (c) and (d) requiring familiarity with the large data set, only a small minority Most students successfully answered part (a) with a correct interpretation of a lack of correlation shown in Figure 1.

Common mistakes

• no (or weak) correlation
• positive/negative ignored

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

Examiner's comment

This question was the second most accessible question on the paper thanks to part (a) and (b), but due to parts (c) and (d) requiring familiarity with the large data set, only a small minority Part (b) was also answered well with students answering in the context of temperature and pressure. There were some students who did not earn the mark as they only wrote negative correlation and others that described a negative correlation as one quantity decreasing as the other decreased. It was notable that many students are still unfamiliar with the large data set and the general standard in parts (c) and (d) was poor. Most students guessed an answer to part (c) but very

Common mistakes

• must interpret in context: as pressure increases temperature decreases
• "negative correlation" alone or "inversely proportional" scores 0

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

Examiner's comment

but due to parts (c) and (d) requiring familiarity with the large data set, only a small minority of students scored full marks here. Most students successfully answered part (a) with a correct interpretation of a lack of standard in parts (c) and (d) was poor. Most students guessed an answer to part (c) but very few were able to give an answer in the correct range. In part (d), the students often described

Common mistakes

• interpolation needs a pressure within the LDS range for Beijing 2015, 990 to 1040 hPa
• most students guessed

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

Examiner's comment

but due to parts (c) and (d) requiring familiarity with the large data set, only a small minority of students scored full marks here. Most students successfully answered part (a) with a correct interpretation of a lack of standard in parts (c) and (d) was poor. Most students guessed an answer to part (c) but very few were able to give an answer in the correct range. In part (d), the students often described the location of Beijing as a factor or repeated an answer given in previous examination series rather than comment on the Beaufort Conversion being qualitative data. It was clear the students who had studied the large data set as their answers had a concise explanation.

Common mistakes

• Daily Mean Wind Speed (Beaufort Conversion) is qualitative/categorical (given in words)
• "not continuous" alone not accepted
• location of Beijing is not a reason

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

Examiner's comment

This was the second most demanding question on the paper. In this question there was some confusion between the number of students who study Art and the number of students who study Art but not Music. This often led to incorrect Venn diagrams and subsequent errors in both parts of the question. In part (a), students often found that at most 20 students studied Music but not Art, but then went no further. Very few were able to find both required end-points.

Common mistakes

• Venn reasoning: smallest x = 12 (all in both), largest 12 + (45 − 25) = 32
• answer 12 ≤ x ≤ 32 with "and" (strict or "or" loses the A mark)
• confusion between Art and Art-but-not-Music

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

Examiner's comment

This was the second most demanding question on the paper. In this question there was some confusion between the number of students who study Art and the number of students who study Art but not Music. This often led to incorrect Venn diagrams and subsequent errors in both parts of the question. The rule for independence is given to students on the formulae booklet, but many quoted it in terms of A and B rather than the events A and M as required in part (b) of this question. Students making little progress often tended to use numbers rather than probabilities throughout. Only a small minority arrived at the correct result of x = 21.6, but some then concluded that A and M were independent as 21.6 was in the range, ignoring the fact that x had to be an integer in this context.

Common mistakes

• P(A) × P(M) = P(A and M) written with A and M (not A and B)
• P(M) = 0.48 gives x = 21.6 (or 9.6 Music only)
• not an integer, so not possible
• working with numbers instead of probabilities
• concluding independent because 21.6 is in the range from (a)

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

Examiner's comment

There were some good attempts seen at this question with most students able to display some knowledge of hypothesis testing. Part (a) was very well answered with most students scoring this mark for a correct alternative hypothesis using the correct notation. A few students did not appreciate that this was a two- tailed test and gave a one-tailed alternative hypothesis.

Common mistakes

• two-tailed: H1: p ≠ 0.25 in terms of p
• some gave a one-tailed H1

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

Examiner's comment

There were some good attempts seen at this question with most students able to display some knowledge of hypothesis testing. In part (b), when attempted, most students acknowledged that it was a Binomial distribution and many were able to find the probability associated with at least one of the tails (usually the lower one). However, some did not take notice in the question that the probability in each tail was required to be as close to 2.5% as possible, and hence found an incorrect upper critical region. Marks were sometimes lost for failing to state the probability associated with each tail (particularly the upper tail) and for incorrectly identifying the upper critical region, usually opting for X >= 20. There are a significant number of students writing the critical region as a probability. Many students with an incorrect critical region were still able to earn the mark in

Common mistakes

• X ~ B(50, 0.25)
• P(X ≤ 6) = 0.0194 and P(X ≥ 19) = 0.0287 must both be stated
• CR X ≤ 6 or X ≥ 19 (X ≥ 20 common error)
• point probabilities such as P(X = 6) are not tail probabilities
• CR written as a probability

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

Examiner's comment

There were some good attempts seen at this question with most students able to display some knowledge of hypothesis testing. part (c) using the correct probabilities for their tails.

Common mistakes

• actual significance level = sum of the two tail probabilities = 0.048 (ft their CR)

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

Examiner's comment

There were some good attempts seen at this question with most students able to display some knowledge of hypothesis testing. Part (d) was answered well by many students, but, as usual, some did not interpret the conclusion in context. Some ignored the hypothesis test and critical regions altogether and simply compared 10 as a percentage of 50 with 25% to suggest a change in probability.

Common mistakes

• 10 is not in the CR: do not reject H0
• conclusion in context (proportion with the allergy, Rylan's belief not supported)
• comparing 10/50 = 20% with 25% instead of using the CR

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

Examiner's comment

The final question on the paper, as expected, proved to be the most demanding as many students were unable to make any progress at all here. Part (a) was the most discriminating with many students leaving this part blank. Students often thought that X followed a binomial distribution. Some tried to work out individual probabilities for each letter rather than focusing on the number of A’s that could be selected. A large number of tree diagrams were seen but some could not get from the diagram to a probability distribution as they were unable to give a probability for each of the different values of X. Students who correctly worked out the probability of obtaining 1 or 2 A's sometimes did not recognise that the answer needed multiplying by 3.

Common mistakes

• X takes only 0, 1, 2 (two A's in VARIANCE)
• without replacement: P(0) = 6/8 × 5/7 × 4/6 = 5/14, P(1) = 15/28, P(2) = 3/28
• multiply by 3 for the arrangements (often forgotten)
• X is not binomial

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

Examiner's comment

The final question on the paper, as expected, proved to be the most demanding as many students were unable to make any progress at all here. Part (b) saw a higher standard. Students were able to work out that a binomial distribution needed to be used and many were able to give the correct value of n and p. One common mistake involved choosing 9 trials rather than 10 as there are 9 letters in DEVIATION.

Common mistakes

• with replacement so binomial B(10, 1/9) (one E in DEVIATION's 9 letters)
• P(J ≥ 4) = 1 − P(J ≤ 3) = 0.0184
• common error n = 9 trials