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Averages, spread and comparing distributions worksheet — GCSE Foundation

Fifteen questions on "averages, spread and comparing distributions" — DfE statement S4. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.

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Averages, spread and comparing distributions worksheet — GCSE Foundation

Maths · GCSE Foundation · S4 · Averages, spread and comparing distributions · Non-calculator · 15 questions

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1.Two classes sat the same test. The 30 pupils in Class A had a mean mark of 72. The 20 pupils in Class B had a mean mark of 82. Work out the mean mark of all 50 pupils.
(a)3800 marks
(b)78 marks
(c)76 marks
(d)77 marks
2.Seven pupils were asked how many books they had read last month. Their answers were 3, 5, 5, 7, 8, 5, 3. Write down the mode.
(a)7
(b)8
(c)5
(d)3
3.Noah wrote down the numbers 9, 3, 7, 1, 5 and said, “The median is 7, because 7 is in the middle of my list.” Is Noah right? Give a reason for your answer.
(a)Yes — the mean of the five numbers is 7, so the median is 7 as well.
(b)No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5.
(c)No — in order the numbers are 1, 3, 5, 7, 9, so the median is 3.
(d)Yes — 7 is the third of the five numbers he wrote, so it is the median.
4.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
(a)62.5
(b)75
(c)37.5
(d)50
5.Harry counted the coins he found on each of six days: 5, 7, 8, 9, 11, 30. Write down the outlier.
(a)30
(b)8.5
(c)11
(d)5
6.The mean of 5 numbers is 8. Work out the total of the 5 numbers.
(a)40
(b)1.6
(c)13
(d)8
7.A netball team scored 50, 83 and 68 points in three matches. Work out the mean number of points scored.
(a)68
(b)33
(c)67
(d)100.5
8.A café sold 10 sandwiches on each of four days: 10, 10, 10, 10. Work out the range of the numbers sold.
(a)20
(b)40
(c)10
(d)0
9.The masses of eight school bags, in kilograms, are 3, 4, 4, 5, 6, 7, 8 and 11. Work out the median mass.
(a)5 kg
(b)5.5 kg
(c)8 kg
(d)4 kg
10.Work out the median of these six numbers: 13, 21, 22, 36, 37, 47
(a)22
(b)36
(c)34
(d)29
11.Grace asked 12 children how many brothers and sisters they have. Her results, in order, were 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 6. Work out the median number of brothers and sisters.
(a)1.5
(b)1
(c)6
(d)2
12.Five pupils spent these numbers of minutes on their homework: 50, 65, 55, 90, 60. Work out the median time.
(a)64 minutes
(b)40 minutes
(c)60 minutes
(d)70 minutes
13.The marks scored by four pupils in a quiz were 8, 8, 8, 8. Work out the mean, the median and the mode of these marks.
(a)mean = 2, median = 8, mode = 8
(b)mean = 8, median = 8, mode = 4
(c)mean = 8, median = 8, mode = 8
(d)mean = 32, median = 8, mode = 8
14.Two classes sat the same maths test, both marked out of 100. Class A had a mean mark of 70 and a range of 30 marks. Class B had a mean mark of 70 and a range of 10 marks. Compare the marks of the two classes.
(a)The means are equal, and Class B's marks are the more consistent because its range is smaller.
(b)Class A's mean mark is higher, and Class A's marks are also the more consistent.
(c)The means are equal, and Class A's marks are the more consistent because its range is larger.
(d)Class B's mean mark is higher, and its smaller range makes its marks the more consistent.
15.A shop sold seven pairs of shoes in these sizes: 4, 4, 5, 6, 6, 6, 9. Write down the modal size.
(a)3
(b)9
(c)4
(d)6

Answer key

(c) 76 marks — Method: a mean of means only works when the groups are the same size, so rebuild each class's total mark, add the totals and divide by the number of pupils altogether. Working: Class A scored 30 × 72 = 2160 marks and Class B scored 20 × 82 = 1640 marks, giving 2160 + 1640 = 3800 marks between 50 pupils, so the overall mean is 3800 ÷ 50 = 76 marks. Answer: 76 marks. The distractors: 77 marks comes from averaging the two class means, (72 + 82) ÷ 2, which ignores the different class sizes; 78 marks comes from attaching each mean to the other class's size, (30 × 82 + 20 × 72) ÷ 50; 3800 marks comes from stopping at the combined total and never dividing by 50.
(c) 5 — Method: the mode is the value that occurs most often, so count how many times each different value appears and compare the counts. Working: 3 appears twice, 5 appears three times, 7 appears once and 8 appears once, so the highest frequency is three and the value carrying it is 5. Answer: 5. The distractors: 3 comes from writing down the frequency of the most common answer instead of the answer itself; 7 comes from taking the middle number of the list as it was written, which applies the median without ordering the data and without answering the question asked; 8 comes from picking the largest value, which confuses the mode with the maximum.
(b) No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. — Method: the median is the middle value of the data in order of size, so the data must be sorted before any position is read off. Working: Noah's list 9, 3, 7, 1, 5 is not in order; sorted it becomes 1, 3, 5, 7, 9, and with 5 values the middle position is the third, which now holds 5 rather than 7. Noah has read the third value of the unsorted list. Answer: no — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. The distractors: the reply giving 3 as the median sorts the data correctly but then reads the value in the second place instead of the third; the reply that 7 is the third number he wrote accepts a position in the unsorted list, which is exactly the mistake the question is about; the reply using the mean claims a value of 7 for it, but the mean is 25 ÷ 5 = 5, so that reasoning is false as well.
(a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
(a) 30 — Method: an outlier is a value that lies far away from the pattern set by the rest of the data, so compare each value with the group the others form. Working: five of the counts, 5, 7, 8, 9 and 11, lie within 6 of one another and the steps between them are 2, 1, 1 and 2; the remaining count of 30 is 19 above the nearest of them, so it is the value that does not belong to the pattern. Answer: 30. The distractors: 5 comes from picking the smallest value, on the idea that the odd one out must be at the bottom of the list; 11 comes from ordering the data and stopping one value short, taking the largest of the counts that sit close together; 8.5 comes from working out the median, (8 + 9) ÷ 2, and giving a measure of centre where a value standing apart was asked for.
(a) 40 — Method: the mean is the total divided by how many values there are, so rearranging gives total = mean × number of values. Working: the mean is 8 and there are 5 numbers, so the total is 8 × 5 = 40. Answer: 40, and checking, 40 ÷ 5 = 8, which is the mean given. The distractors: 13 comes from adding the mean and the count, 8 + 5, instead of multiplying them; 1.6 comes from dividing the mean by the count, 8 ÷ 5, which reverses the relationship; 8 comes from quoting the mean itself as the total, which is only true when there is a single number.
(c) 67 — Method: the mean is the total of the values divided by how many values there are, so add first and divide second. Working: the total is 50 + 83 + 68 = 201 points and three matches were played, so the mean is 201 ÷ 3 = 67 points. Answer: 67. The distractors: 68 comes from writing down the median, the middle value of 50, 68, 83, instead of the mean; 33 comes from working out the range, 83 − 50, which measures spread and not centre; 100.5 comes from dividing the total by 2 instead of by the 3 matches played.
(d) 0 — Method: the range is the largest value minus the smallest value, whatever those two values turn out to be. Working: every value is 10, so the largest value is 10 and the smallest value is 10 as well, and the range is 10 − 10 = 0. Answer: 0 — a range of nothing says the data do not vary at all. The distractors: 10 comes from writing down the repeated value itself instead of the difference between the extremes; 20 comes from adding the largest and the smallest, 10 + 10, instead of subtracting; 40 comes from adding all four values, which gives the total sold and not a measure of spread.
(b) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.
(d) 29 — Method: with an even number of values there is no single middle value, so the median is the mean of the two values either side of the middle. Working: the six numbers are already in order and 6 ÷ 2 = 3, so the middle pair are the third and fourth values, 22 and 36; their mean is (22 + 36) ÷ 2 = 58 ÷ 2 = 29. Answer: 29, which lies between the two middle values as a median of an even data set must. The distractors: 22 comes from taking the lower of the two middle values and stopping there instead of averaging the pair; 36 comes from taking the larger value of that pair because it sits just past the halfway point of the list; 34 comes from working out the range, 47 − 13, instead of a measure of centre.
(a) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
(c) 60 minutes — Method: the median is the middle value once the data have been put in order of size, so the list must be sorted before any position is read. Working: in order the times are 50, 55, 60, 65, 90 minutes; there are 5 values, so the middle position is the third and the time sitting there is 60 minutes. Answer: 60 minutes. The distractors: 64 minutes comes from working out the mean, 320 ÷ 5, instead of the median; 70 minutes comes from taking the time halfway between the shortest and the longest, (50 + 90) ÷ 2; 40 minutes comes from working out the range, 90 − 50, which measures spread rather than centre.
(c) mean = 8, median = 8, mode = 8 — Method: work out each measure separately — the mean is the total divided by how many values there are, the median is the middle value once the data are in order, and the mode is the value that occurs most often. Working: the total is 8 + 8 + 8 + 8 = 32 and there are 4 marks, so the mean is 32 ÷ 4 = 8; in order the marks read 8, 8, 8, 8, and the mean of the middle pair is (8 + 8) ÷ 2 = 8; the value 8 occurs 4 times and no other value occurs at all, so the mode is 8. Answer: mean = 8, median = 8, mode = 8 — when every value in a data set is the same, all three measures of central tendency take that value. The distractors: a mean of 32 comes from stopping at the total and never dividing by 4; a mode of 4 comes from writing down how many times 8 occurs instead of the value that occurs; a mean of 2 comes from dividing a single value, 8, by the 4 marks instead of dividing the total by 4.
(a) The means are equal, and Class B's marks are the more consistent because its range is smaller. — Method: comparing two distributions needs two things — a measure of average and a measure of spread — and each must be put into the context of the question. Working: both classes have a mean mark of 70, so on average the two classes scored the same; the range measures spread, and Class A's range of 30 marks is three times Class B's range of 10 marks, so Class B's marks sit closely around the mean while Class A's are far more spread out. Answer: the means are equal, and Class B's marks are the more consistent because its range is smaller. The distractors: the reply crediting Class A with more consistency reverses the meaning of the range, treating a larger range as tighter data when a larger range means more spread; the reply that Class A's mean mark is higher compares the wrong pair of figures, reading the range of 30 as an average; the reply that Class B's mean mark is higher reads the spread correctly but its claim about the means is false, since both means are 70.
(d) 6 — Method: the mode, or modal value, is the value that occurs most often in the data set, and it is a value from the data rather than a count. Working: size 4 occurs twice, size 5 occurs once, size 6 occurs three times and size 9 occurs once, so the highest frequency is three and the size it belongs to is 6. Answer: 6. The distractors: 3 comes from writing down the frequency of the most common size instead of the size itself; 9 comes from picking the largest size in the list, which confuses the mode with the maximum; 4 comes from stopping at the first size that repeats rather than checking which size repeats most often.

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