Write down the modal class.

Write down the mid interval value for the class 6 ≤ x < 10 .

Calculate an estimate of the mean annual snowfall.

Find the number of years for which the annual snowfall was at least 18 cm.

14 ≤ x < 18     (A1)     (C1)

[1 mark]

8     (A1)     (C1)

[1 mark]

\(\frac{{{\text{4}} \times {\text{30 + 8}} \times {\text{26 + 12}} \times {\text{29 + 16}} \times {\text{32 + 20}} \times {\text{18 + 24}} \times {\text{27 + 28}} \times {\text{14}}}}{{176}}\)     (M1)

Notes: Award (M1) for an attempt to substitute their mid-interval values (consistent with their answer to part (b)) into the formula for the mean. Award (M1) where a table is constructed with their (consistent) mid-interval values listed along with the frequencies.

= 14.7 (cm) (14.7045…)     (A1)(ft)     (C2)

Notes: Follow through from their answer to part (b). If a final incorrect answer that is consistent with their (b) is given award (M1)(A1)(ft) even if no working is seen.

[2 marks]

18 + 27 + 14     (M1)

Note: Award (M1) for adding 18, 27 and 14.

= 59     (A1)     (C2)

[2 marks]

Part (a) was generally well done but, in part (b), writing down the mid-interval value of a class proved difficult for some candidates and many incorrect answers of 7.5 were seen.

Part (a) was generally well done but, in part (b), writing down the mid-interval value of a class proved difficult for some candidates and many incorrect answers of 7.5 were seen.

Popular, but erroneous answers, seen in part (c) were 15.5 and 16. These seemed to be as a result of adding their mid-class values together and dividing by 7 rather than the total of the frequencies.

There was much confusion over the meaning of the phrase ‘at least’ in part (d) and, as a consequence, there were as many wrong answers of 117 (30 + 26 + 29 + 32) seen as there were correct answers.