S4.3 Choosing a distribution

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

Examiner's comment

This question also proved to be quite challenging. often made and a divisor of 1600 not 1800 was often seen. Part (b) though was not answered so well. Some did explain that the histogram was not symmetric or was skewed and so a normal distribution wouldn’t be suitable. Some focussed on the continuous nature of the variable and thought it was therefore suitable whilst others thought that T was discrete and therefore a normal wasn’t suitable. The question wanted reference to the histogram and there was no expectation or requirement to find probabilities from the normal distribution nor compare the mean of the data with 14.9 If further calculations were expected these would be asked for and reflected in the mark allocation. Many students were able to use integration by parts correctly to establish the result in part

Common mistakes

• histogram is skewed/not symmetric but the normal is symmetric
• arguments about T being continuous or discrete score B0
• no calculation expected

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

Examiner's comment

There was a wide variety of responses to part (d). Successful students often stated one or more of the conditions of the Binomial Distribution, with reasons why this distribution was not suitable in this case. e.g. • the probability needs to be independent but the probability of rainfall is not independent as it often depends on the weather from the day before. • the probability of rain needs to be constant but the probability was taken from May to October and summer is June to September and therefore the probability is likely to be higher than in the summer when there is less rain Errors included • referring to independence without context • commenting on the great storm of 1987 or global warming • stating that the data is out of date • the sample size is too small • some values are trace values and/or rainfall is influenced by other factors.

Common mistakes

• probability of a dry day unlikely to be constant over the year (estimate is not from summer) or days not independent (in context)
• comments on 1987 storm, sample size or trace values score 0

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

Examiner's comment

It was disappointing that over 20% of candidates scored no marks here as there were some straightforward marks available to those who had studied this part of the

Common mistakes

• a significant (non-negligible) probability of T < 0, which is impossible
• just "can be negative" is not enough as this is true of any normal model

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

Examiner's comment

candidates. Parts (b) and (e) were a problem for those less familiar with the large data set. Part (e) was very poorly done with only a handful of candidates scoring marks here. Depth of familiarity was lacking by many candidates. Some misunderstood the question entirely and cited the conditions for a Normal being a suitable approximation to a Binomial. Others thought this part was related to the variable under study and suggested the Normal was not suitable for modelling air temperature for a variety of reasons including the skew on the box plot. Those who did suggest two other variables from the LDS rarely gave acceptable reasons for their choice; some gave no reasons at all even when their choice was a suitable one. The use of the Beaufort scale for wind speed was mentioned but it was wrongly said to be discrete with only a few saying that it was non-numeric. Wind direction which was non-numeric was quite often seen but was not an acceptable answer. Some other wrong reasons were: the data is non- uniform, the values are too variable; it is not possible to have values below 0. The most commonly suggested variable was cloud cover, presumably because it was used as the basis for Q4 and possibly because it appeared in last summer’s 8MA02

Common mistakes

• rainfall (skewed, many days of 0), wind speed Beaufort (non-numeric), pressure, date (uniform)
• "discrete" or "not continuous" not accepted
• wind direction not accepted
• normal-approximation-to-binomial conditions are irrelevant

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

Examiner's comment

This question demonstrated that most candidates are confident in carrying out statistical calculations, particularly probabilities, but only the most able are confident in interpreting in Part (c) was not particularly well attempted by candidates with many leaving it out completely. Where there was a clear attempt those failing to score the mark usually only compared one set of the values, generally the 51.7 and the 52. Most who attempted part (d) scored the mark.

Common mistakes

• compare 0.717 with 0.703 and 51.7 with 52 (probability with probability, days with days) and conclude the model is supported
• comparing only one pair

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

Examiner's comment

This question demonstrated that most candidates are confident in carrying out statistical calculations, particularly probabilities, but only the most able are confident in interpreting in Part (e) was the least successfully answered part of the question with lots of blanks seen. Where candidates scored the 2 marks it was usually through stating that if there had been cloud cover the day before there was a higher chance of cloud cover the next day. Those going completely down the wrong route failed to realise that the data was from the same source and were mostly commenting on the size of the sample rather than what the statistics already demonstrated. In general there was far too much tendency to describe vague contextual knowledge rather than use the figures found previously. Of those who did try and compare figures an error often seen was to compare the result in (d) to 0.76. The failure to focus on the fact that the model was

Common mistakes

• 0.821 is higher than 0.717 – high cloud cover is more likely after a day of high cover, so independence fails and the binomial model may not be suitable
• comparing with 0.76 or commenting on sample size is wrong

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

Examiner's comment

part (c) but responses to part (d) were poor. This question required students to suggest a refinement to Helen’s model and most answers lacked the detail required for the mark.

Common mistakes

• compare 0.315 with 4/9: observed proportion is lower, so the uniform model is not good
• comment must be based on their (b)

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

Examiner's comment

part (c) but responses to part (d) were poor. This question required students to suggest a refinement to Helen’s model and most answers lacked the detail required for the mark.

Common mistakes

• just saying "not uniform" is B0
• refinement must use variation by month or location (e.g. a non-uniform distribution with probabilities from frequencies)
• a continuous model such as normal is B0
• most answers lacked detail

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

Examiner's comment

In part (a), we would expect answers to use the context and although many stated that Peta assumes that the probability of hitting the target is constant most did not mention that the throws of the dart had to be independent too, many simply referring to just “trials”.

Common mistakes

• probability of hitting is constant and throws are independent, in context
• "trials are independent" is B0
• many omitted independence