Question Working Answer Mark AO Notes Grade 1 a b × 50 000 28 000 × 0.9 = 25 200 28 000 would vote for Party A. This is over 50% of voters so Party A should be confident of winning. M1 M1 A1 2 M1 for multiplication of total population by proportion of sample M1 for 28 000 × 0.9 A1 for explanation that this represents over half the whole population and so a majority is likely B 3 2 × 1260 = 315 315 students are in the sample. M1 A1 2 M1 for multiplication of total population by proportion of sample A1 cao B 2 3 At four different points in the concert, entrances or bars, each point to conduct the survey on approximately 100 people each, trying to ask as many males as females. B1 B1 2 3 B1 for explanation of general method of sampling B1 for specific example related to the concert crowd B 2
6 a b c d 1 + 12 + 24 + 15 + 13 + 9 + 5 = 79 Total number of minutes = sum of (midpoint × frequency) for each class = 2754 Mean time = = 34.86… minutes Estimated mean = 35 minutes Modal time = 21–30 minutes= 40 The median is the 40th person. The 40th person is in the 31–40 group. Median time is 31–40 minutes For the group who waited over one hour (61–70), there were 5 patients. So 74 patients were seen in one hour or less. = 0.9367…. So 93.7% of patients saw a doctor within one hour. 79 patients Estimate mean waiting time = 35 minutes The hospital would use the modal time. 93.7% of patients saw a doctor within one hour. M1 M1 M1 M1 B1 M1 M1 A1 M1 A1 3.1 M1 for addition of frequencies M1 for midpoints × frequencies M1 for total minutes M1 for division of total minutes by total patients (ft) B1 for appropriate rounding to whole minutes M1 for mode M1 for median A1 for correct, shortest, average waiting time M1 for calculation of or 1 – () as a percentage A1 cao B 10
5 14 Either Harold, as he had bigger tomatoes, or Connie, as she had more tomatoes (depending on if you want lots of tomatoes or large tomatoes!). M1 A1 2 M1 for choice of one person with justification A1 for further consideration of second person M 2 15 a bi ii The statement: ‘There is strong evidence to support my hypothesis’ is more precise. In reality it is usually not possible to prove a hypothesis just to gather evidence to support it. Because you may not have gathered enough evidence. Because you haven’t gathered enough evidence or you’re using a ‘null’ hypothesis. M1 A1 A1 2 M1 for selection of ‘There is strong evidence to support my hypothesis’ with explanation about practicalities of proving an hypothesis A1 for explanation of having sufficiently large sample / evidence A1 for giving same reason / an awareness of the practical difficulties of getting all other variables equal or unchanged so that they do not affect the success or otherwise of your evidence gathering. oe M 3 16 a b c Valid comparative statement such as: ‘The means are very similar but there is evidence that the broadsheets use a broader range of length of words.’ No, it provides some evidence, but only assuming length of words is a suitable measure of difficulty. The length of words may not be the thing that readers find makes a paper hard to read. For example: How long did it take to read the sample or a specific question to challenge the readers’ understanding of the sample.’ Could also look at the modal word length to see if there is evidence to support the B1 B1 B1 B1 for comparative statement oe B1 for explanation that this is insufficient evidence to answer the question oe B1 for a suitable further question oe H
19 Students’ diagrams, measures and reports will vary. B1 B1 M1 M1 A1 3 B1 for report that includes an awareness of change over time and which wines are increasing and which are decreasing in popularity B1 for appropriate diagram / graph oe M1 for calculation of ranges to show which wines are most popular oe. M1 for calculation of mean, median or mode to show average number of cases of wine purchased purchased oe A1 for interpretation of calculations within the report H 5 20 Estimated mean = = 27.6… minutes From the cumulative frequency (cf) graph (some variance depending on students’ reading of unmarked intervals on graph): Estimate: 28 minutes (to the nearest minute). M1 M1 A1 A1 2 3 M1 for identification of frequencies from cf graph M1 for identification of mid-points of classes from cf graph A1 for table to show data derived from cf graph oe A1 for their estimated mean (ft) Allow range of 26–29 minutes H 4 21 Find the top 15% on the cumulative frequency scale, read along to the graph and read down to the marks. The mark seen will be the minimum mark needed for this top grade. M1 A1 2 M1 for explanation of use of graph A1 for explanation of where to find the minimum grade H 2
29 a b Suitable statements about the distributions with justification for why the representations are from different distributions. Distribution 1 is for Africa, distribution 2 is for Europe, plus observations such as: there is a peak at 21; distribution 1 is positively skewed; there is a more constant frequency across the range for distribution 2 with more women marrying older in distribution 2. B1 B1 2 B1 for comparison statements such as D1 has data for ages 18–11, which D2 does not; D2 has data for ages 30–32, which D1 does not; D1 has a positive skew so it has a low mode with a long right tail and D2 has a slight negative skew, so it has a higher mode but more consistency across a greater range oe B1 for observation of skewness and students’ comments on and perception of possible reasons such as: women in Africa tend to marry at a younger age; European women have better health care or careers and so marry later; there is more gender equality and independent women in Europe oe H 2 30 a b c Examples such as: positive skew – income distribution negative skew – age at death in developed countries. The box will be shifted to be in line with the bulk of the distribution. It will veer right for negative skew, left for positive and central for no skew. Suitable box plot for each distribution. B1 B1 B1 B1 B1 B1 2 B1 for positive skew context where there is a long tail to the right (higher values) and the bulk of the distribution is to the left (lower values) B1 for negative skew context where there is a long tail to the left (lower values) and the bulk of the distribution is to the right (higher values) B1 for description of negative skew that includes a comment on the distribution being focused to the right (lower values) oe B1 for suitable positive skew box plot B1 for suitable no skew box plot B1 for suitable negative skew box plot Positive skew (long right tail, median to the left) H 6