June 2024: AS AS Paper 2 Section A: Statistics

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

Examiner's comment

This question proved to be a successful start to the paper for most students, with many scoring full marks. Those who struggled did not recognise that 25% of the players' heights represented the lower quartile. They were still able to draw the interquartile range to the correct distance. Most students were able to correctly draw the whiskers for the lower and upper values, but there were a small minority who have no idea what a box and whisker diagram should look like, and some just plotted points or simply vertical drew lines.

Common mistakes

• "25% below 1.81" means Q1 = 1.81
• Q3 = 1.81 + 0.11 = 1.92
• whiskers from 1.72 to 1.72 + 0.28 = 2.00
• median 1.85 inside the box (not equal to Q3)
• some plotted points or vertical lines only

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

Examiner's comment

This question was generally not well attempted demonstrating that students are still not completely familiar with the large data set. In part (a) many students appeared to be aware of Camborne's location in the South-West of the UK but could not relate this to the prevailing winds in that area. Some students were able to guess that A represented East but were unable to provide a reason. Many students did not appreciate that wind direction is the direction that wind blows from, rather than to.

Common mistakes

• A is East: the least common direction, since prevailing UK/Camborne winds come from the south-west (west/south)
• wind direction is where the wind blows from, not to
• a reason is needed

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

Examiner's comment

This question was generally not well attempted demonstrating that students are still not In part (b), many students appreciated that the value was an outlier and should be excluded but were unable to score the marks here without justification. Those who understood that the maximum that the angle could be 360, tended to score well here. Some students also thought that 999 represented a frequency rather than an angle.

Common mistakes

• wind direction runs from 0 to 360 degrees so 999 is not a valid value (it is an error/missing-data code, not an outlier or a frequency)
• remove/ignore it
• a reason using 360 is needed before the action

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

Examiner's comment

Part (a) was well attempted and almost all students realised that the two missing frequencies should sum to 32.

Common mistakes

• deduce the frequency density scale from the 3–4 bar (frequency 13)
• 2–3 has frequency 25 and 4–6 has 112 − (64 + 25 + 13 + 3) = 7
• the two missing frequencies sum to 32

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

Examiner's comment

In part (b), the most usual mistake was to use frequency rather than frequency density for the heights of the 4 – 6 and 6 – 8 bars, drawing them at 7 small squares and 3 small squares respectively. Also, the first bar was often seen drawn from 0 – 2 instead of 1 – 2 often with an incorrect height.

Common mistakes

• bar heights are frequency densities: 4–6 bar height 7/2 = 3.5 and 6–8 bar height 1.5
• drawing heights as frequencies (7 and 3)
• first bar must run from 1 to 2, not 0 to 2

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

Examiner's comment

Part (c) was not well done and quite often no attempt was made. Some students substituted one or more values of x, setting the result equal to the frequency (or frequency density). When integration was used, it was often done well. It was, however, surprising to see that even when the correct integration was equated to 112, many were unable to accurately solve the equation. Many students made slips with the negative signs, and some were unable to solve the equation, multiplying one side by 8 to get rid of the fraction but failing to multiply 112 by 8 at the same time.

Common mistakes

• area under the frequency density curve is the total frequency: ∫ from 1 to 8 of k/x² dx = 112
• (7/8)k = 112 gives k = 128
• substituting x values instead of integrating
• sign slips with −k/x

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

Examiner's comment

Part (a) was well answered with the majority of students gaining this mark, the only issues occurring with rounding or calculating P(𝑋 <= 10) rather than P(𝑋 = 10). Part (a)(ii) was less well attempted with many students unsure of the difference between <= and <. The most common error seen was to evaluate P(𝑋 <= 14) – P(𝑋 >= 12) although many other incorrect expressions were seen. Often when students worked out P(𝑋= 12) + P(𝑋= 13) + P(𝑋= 14), they attained full marks. For those who attempted part (b), this was answered much better than in previous series and

Common mistakes

• (i) P(X = 10) = 0.154, not P(X ⩽ 10)
• (ii) P(12 ⩽ X < 15) = P(X ⩽ 14) − P(X ⩽ 11) = 0.179
• common error P(X ⩽ 14) − P(X ⩽ 12)

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

Examiner's comment

For those who attempted part (b), this was answered much better than in previous series and students are becoming more familiar with the structure of these questions, writing the hypotheses in terms of p and the distribution clearly. Sometimes the probability statements were not clearly stated, with students writing down the values that they had inputted into their calculators. It was quite common to see P(𝑋 < 3) or P(𝑋 = 3) rather than P(𝑋 <= 3) calculated. The conclusion was not always given in context. When the context was given, it was usually correct, although missing out the word “proportion” or its equivalent meant the mark was lost. Some students chose to use a critical region approach rather than a p-value approach, but they often did not make this clear and some were then unsure about whether 3 was inside or outside the critical region and what this implied for the acceptance or otherwise of H0 Part (d) was frequently omitted or the value p = 0.12 given. Those who used the critical region approach rarely got this mark. However, students are becoming more aware of what the p-value represents.

Common mistakes

• H₀: p = 0.12, H₁: p < 0.12 (in terms of p)
• D ~ B(60, 0.12)
• P(D ⩽ 3) = 0.0601 > 0.05 (not P(D < 3) or P(D = 3))
• do not reject H₀: insufficient evidence that the proportion of defective items has decreased
• conclusion in context with "proportion"
• CR approach must be made clear (CR D ⩽ 2)

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

Examiner's comment

This question was generally well attempted, but there are still a significant number of students who show no understanding of hypothesis testing. Some students truncated the values seen on their calculators rather than giving them to an appropriate degree of accuracy.

Common mistakes

• p-value = 0.06 (the probability P(D ⩽ 3) from (b))
• giving 0.12 or the CR probability 0.0196 is wrong
• often omitted

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

Examiner's comment

Part (a) and part (b) was generally done well by the vast majority of students. A large number of students overcomplicated their workings by calculating probabilities for all 9 possibilities rather than simply calculating P(𝑛𝑜𝑡 7) x P(𝑛𝑜𝑡 7) x P(7). Many students still do not understand the concept of a probability distribution and many attempts were made to calculate a single probability, usually from a binomial distribution, in

Common mistakes

• probabilities sum to 1: 2q = 0.3, q = 0.15

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

Examiner's comment

Part (a) and part (b) was generally done well by the vast majority of students. A large number of students overcomplicated their workings by calculating probabilities for all 9 possibilities rather than simply calculating P(𝑛𝑜𝑡 7) x P(𝑛𝑜𝑡 7) x P(7). Many students still do not understand the concept of a probability distribution and many attempts were made to calculate a single probability, usually from a binomial distribution, in

Common mistakes

• stops after the 3rd spin means not 7, not 7, then 7: 0.8 × 0.8 × 0.2 = 0.128
• listing all nine sequences is unnecessary

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 and parts (c) and (d) were only completed by the most able students. part (c). Some students were able to make a good attempt at the probability distribution, but a large number failed to correctly calculate P(𝑆= 4) as they didn't account for the spinner failing to land on a 7 four times. Most students left out part (d) as they were unable to interpret what this probability

Common mistakes

• S takes values 1, 2, 3, 4 only
• P(S = 1) = 0.2, P(S = 2) = 0.16, P(S = 3) = 0.128, P(S = 4) = 0.8³ = 0.512 (she stops after 4 spins whether or not she gets a 7)
• P(S = 4) = 0.8³ × 0.2 alone is the common error
• trying a binomial

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

Examiner's comment

and (d) were only completed by the most able students. Most students left out part (d) as they were unable to interpret what this probability represented. Of those who did try, many complicated probability expressions were seen often leading to an incorrect answer.

Common mistakes

• S > N unless she gets a 7 on the first spin (S = N = 1): 1 − 0.2 = 0.8
• most omitted or overcomplicated this