June 2019: AS AS Paper 2 Section A: Statistics

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

Examiner's comment

This proved to be an accessible start to the paper with most scoring at least 1 mark in part (a) for a partial description of a stratified sample. Very few gained the first mark for clearly describing the method of splitting the population into two groups and labelling these groups. Most realised that a random sample must be taken within each strata and the correct calculation for the number of candidates for each year group was commonly seen.

Common mistakes

• label/number each year group separately
• use random numbers to take a simple random sample within each stratum
• 84/140 × 40 = 24 Year 12s and 16 Year 13s
• the labelling mark was rarely earned

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

Examiner's comment

In part (b), most candidates scored the mark for indicating both ‘increase’ and ‘2.8’. A minority however worked out the number of points scored when s = 0.5. Others just mentioned the ‘increase’ but failed to quantify it appropriately.

Common mistakes

• use the gradient: 0.5 × 5.60 = increase of 2.8 marks (both "increase" and 2.8 needed)
• not the value of p at s = 0.5
• "about 3 marks" is B0

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

Examiner's comment

Part (c) was very well answered and most gained a mark here by coming up with a plausible limitation of the model. The majority referred to ‘other factors affecting performance’ or ‘not specifying a range so candidates could not score less than 26.1 marks or could get infinitely high marks’.

Common mistakes

• e.g. predicts best performance for students who never wake up, only valid within the range of the data, ignores other factors, could exceed the maximum mark
• "might not be correlation" is not accepted

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

Examiner's comment

Only around 10% of candidates achieved full marks on this question, but there were many strong attempts made. Most scored the first mark for identifying that x = 0. The most common error across the question was to ignore the intersections and calculate P(A) x P(C) as 0.1 x 0.39. The majority of candidates scored the final method mark for a correct equation in y and z, demonstrating knowledge of probabilities summing to 1. Surprisingly, of candidates who did set up all of the equations required, many were unable to manipulate the algebra to solve them. Those who confused independent and mutually exclusive events tended to make no progress.

Common mistakes

• B and C mutually exclusive so x = 0
• independence: P(A ∩ C) = P(A) × P(C) with P(A) = 0.1 + y + z and P(C) = 0.39 + z (not 0.1 × 0.39 using only the outer regions)
• probabilities sum to 1 gives y + z = 0.15
• z = 0.13, y = 0.02

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

Examiner's comment

Whilst over half of candidates were able to score at least 3 marks on this question, only 3% went on to achieve full marks. Virtually all candidates struggled to name the distribution in part (a) with the vast majority opting for ‘binomial’. It would appear that the ‘uniform’ distribution is a topic unfamiliar to many candidates.

Common mistakes

• (discrete) uniform distribution
• most wrongly said binomial
• continuous uniform is B0

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

Examiner's comment

Whilst over half of candidates were able to score at least 3 marks on this question, only 3% went on to achieve full marks. Virtually all candidates struggled to name the distribution in Part (b)(i) was well answered as most candidates correctly identified the binomial distribution required here and confidently used 1 – P(X <= 6) to calculate P(X >= 7). Usual errors were seen, notably attempting 1 – P(X <= 7). Whilst there was some good use of calculators seen, a significant amount of candidates are still calculating these probabilities out by hand in full which costs them a large amount of time. Candidates are expected to use their calculators to find binomial probabilities and cumulative probabilities from the binomial distribution. Part (b)(ii) was more discriminating and correct attempts here were rare. The most common mistake was made by those candidates who found P(X <= 7) and subtracted {1 – P(X <= 3)} from it. Those candidates who wrote out the required probabilities in full, i.e. P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7), tended to score full marks.

Common mistakes

• X ~ B(28, 0.2)
• P(X ≥ 7) = 1 − P(X ≤ 6) = 0.322 (not 1 − P(X ≤ 7))
• P(4 ≤ X < 8) = P(X ≤ 7) − P(X ≤ 3) = 0.658
• use calculator cumulative functions rather than summing by hand

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

Examiner's comment

In part (a) most candidates referred to data that was n/a, anomalies or outliers with only a small proportion appreciating that the data set contained trace values for rainfall which needed replacing with numerical values.

Common mistakes

• LDS: rainfall includes "tr" (trace, less than 0.05 mm) which must be converted to a number
• n/a, missing data or outliers are not the answer

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

Examiner's comment

For part (b) most candidates are familiar with using linear interpolation though slips appear often. Errors with endpoints often resulted in multiplication by 5. Other mistakes included 23 in the denominator or 6 on the numerator. For those using an (n + 1) method, it was not uncommon to see 7.5 being used in the numerator instead of 7.75.

Common mistakes

• Q3 is the 138th value (184 × ¾)
• 1 + (138 − 131)/24 × 4 = 2.17
• class width 4, not 5
• (n + 1) method uses 138.75

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

Examiner's comment

In part (c) most candidates were able to use the given information to arrive at the required result. There are still a large number of candidates who do not work to the appropriate degree of accuracy and 5.8 and 5.76 were often seen. Candidates are encouraged to write down all figures from their calculator before rounding to ensure accuracy marks are scored. Those who did not achieve full marks here often had an incorrect denominator or forgot to include the square root.

Common mistakes

• σ = √(7704.1875/184 − (539.75/184)²) = 5.77
• n = 184
• do not forget the square root
• give 3 s.f. (5.8 and 5.76 lose the A mark)

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

Examiner's comment

The underlying assumption in part (d)(i) was generally identified but not always with sufficient clarity to gain the mark. Common errors were: ‘data uniformly distributed’ (failing to mention within each class); the average of each class is the midpoint (failing to specify the mean of each class); most of the data is located at the midpoint (failing to identify that the midpoint is assumed to represent all of the data). It was clear from the responses seen here that many candidates have little or no experience with the large data set. In (d)(ii) quite a number referred to the data being based on weather which is unpredictable and that therefore the previous assumption was unrealistic. Good explanations involving the distribution of the bottom class were rare with very few making reference to the majority of data being either 0 or close to 0. Little progress was made in (d)(iii) as many mentioned irrelevancies such as the great storm or the fact that it was during the summer or that the database did not cover a full year. Many did not realise that they were comparing an estimated mean with an actual mean for the same time span and said the actual mean would be higher because the winter months were excluded.

Common mistakes

• (i) values are spread evenly within each class (not "data uniformly distributed")
• (ii) LDS: most days in the first class have 0 or trace rainfall
• (iii) so the actual mean is smaller than the estimate
• comparing with winter months or storms is irrelevant

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

Examiner's comment

Only the most able candidates on the paper made significant progress through this question on hypothesis testing. In part (a) the first error in writing down the alternative hypothesis was identified and corrected more often than the error in the probability statement. Some candidates lost the marks by identifying where the error was but without correctly identifying what the error was. It was not uncommon for candidates to say that the probability distribution was used rather than the cumulative distribution; likely referring to calculator functions rather than demonstrating a formal understanding of hypothesis testing.

Common mistakes

• H₁ should be p > 0.15 (not ≥)
• the test statistic should be P(X ≥ 8) = 0.0698, not P(X = 8)
• say what the correct version is, not just where the error is

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

Examiner's comment

Only the most able candidates on the paper made significant progress through this question Part (b) was less successfully done and where attempted the mark was not always awarded because insufficient details of the calculation and its implication were given. Despite the question stating the test should be one-tailed in part (c) some candidates found a

Common mistakes

• it does affect the conclusion: P(X ≥ 8) = 0.0698 > 0.05 so H₀ is not rejected
• the probability must be quoted with its comparison

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

Examiner's comment

Only the most able candidates on the paper made significant progress through this question Despite the question stating the test should be one-tailed in part (c) some candidates found a two-tail critical region or went for the left hand tail rather than the right hand one. Where the upper tail was found the critical region was usually correct although some gave a critical value rather than a region or wrote it as a probability i.e. P(X >= 9). Some candidates rewrote Julia’s hypothesis test correctly at this point, not appreciating the difference between this and a critical region approach. Part (d) was usually correctly done by those who were successful in part (c). On some occasions answers were given only to two significant figures. Just over 5% of candidates scored full marks on the final question on this paper.

Common mistakes

• one-tailed upper tail of B(30, 0.15): P(X ≥ 9) = 0.0277 < 0.05, P(X ≥ 8) > 0.05
• CR X ≥ 9 (not a two-tailed or lower tail region, not P(X ≥ 9) as a probability, not just the critical value)

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

Examiner's comment

Only the most able candidates on the paper made significant progress through this question Part (d) was usually correctly done by those who were successful in part (c). On some occasions answers were given only to two significant figures. Just over 5% of candidates scored full marks on the final question on this paper.

Common mistakes

• actual significance level = P(X ≥ 9) = 0.0278 to 3 s.f.