Printable · GCSE Foundation · ages 14-16
Statistics worksheet — GCSE Foundation
Fifteen questions across the statistics statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Statistics worksheet — GCSE Foundation
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- (c) 20 kg ≤ mass < 30 kg — The modal class is the class with the highest frequency. Reading the plotted points, the frequencies are 6, 10, 16, 6 and 2, so the highest frequency is 16, plotted at the midpoint 25. A class of width 10 centred on 25 runs from 25 − 5 = 20 to 25 + 5 = 30, so the modal class is 20 kg ≤ mass < 30 kg. Writing '25 kg' gives only the midpoint, not the class — the modal class is an interval, not a single value. '10 kg ≤ mass < 20 kg' is the class before the peak, centred on 15, which has frequency 10, not the highest. '30 kg ≤ mass < 40 kg' is the class after the peak, centred on 35, which has frequency 6, not the highest.
- (c) 140 — Method: the mean shares the total equally between the items, so add the values and then divide by how many there are. Working: the total is 120 + 200 + 160 + 80 = 560 pages and there are 4 books, so the mean is 560 ÷ 4 = 140 pages. Answer: 140. The distractors: 560 comes from stopping at the total number of pages and never dividing by 4; 280 comes from dividing the total by 2 instead of by the 4 books; 120 comes from working out the range, 200 − 80, which measures spread instead of centre.
- (b) Ethan, 10 seconds — Method: over the same distance the fastest runner is the one who takes the least time, so the smallest time in the table is found first and the name is then read from the same row. Working: the four times are 12 seconds, 15 seconds, 10 seconds and 14 seconds; in order of size these are 10, 12, 14 and 15, so the least time is 10 seconds, and the row holding 10 seconds is the row for Ethan. Answer: Ethan, 10 seconds — the time is in seconds, and a smaller time means a faster runner. The distractors: Grace with 15 seconds comes from taking the largest number in the table to mean the fastest runner, which reverses the relationship between time and speed over a fixed distance; Oliver with 12 seconds comes from writing down the first row of the table without comparing the four times; Ethan with 15 seconds comes from identifying the right runner but then reading the time from a different row of the table.
- (d) It asks two things at once and invites agreement — Method: a survey question is faulty when a reply to it cannot be read as evidence about one single thing, so check how many claims it contains and whether its wording pushes the reader one way. Working: the question joins two separate claims, that learning matters and that more homework should be set, so a reply of yes could mean either of them or both and cannot be counted as evidence about homework; the opening words “Do you agree” also invite agreement instead of leaving the reader free to say no. Answer: it asks two things at once and invites agreement. The distractors: the reply calling it too short mistakes length for clarity, when the fault is that too much has been packed in rather than too little; the reply about long words is false, since every word in the question is an everyday one and the fault lies in what is being asked rather than in the vocabulary used to ask it; the reply that the question is fine takes a yes or no answer as proof that the question works, which is exactly what a double question defeats.
- (a) The median, as the one very large wage does not move it — Method: an average describes a population well when it sits close to most of the values, so compare what each average does when one value lies far from the rest. Working: in order the wages are 420, 440, 460, 480 and 1,500, so the median is the third of the five, £460. The mean uses every wage: 420 + 440 + 460 + 480 + 1,500 = 3,300 and 3,300 ÷ 5 = 660, so the mean is £660. Four of the five people earn less than £660, and the nearest of those four wages is £180 below it, so £660 describes nobody at the garage; £460 sits inside the group of four similar wages. Answer: the median, as the one very large wage does not move it, while that same wage drags the mean £200 above the median. The distractors: saying the median is always larger than the mean is an invented rule, and here the median £460 is smaller than the mean £660; saying the mean is the only average that uses all five wages is true as far as it goes, but using a value and being dragged by it are the same thing when that value is £1,500; saying £660 lies between the smallest and largest wage is true of every mean ever calculated, so it proves nothing about whether this one is typical.
- (a) 35 minutes — Chloe arrives at 09:30, so she needs the first departure time that is at or after 09:30. The bus at 09:25 has already left five minutes before she arrives, so it is missed, not caught; the next one on the timetable is 10:05. From 09:30 to 10:00 is 30 minutes, then a further 5 minutes to 10:05, so the wait is 35 minutes. Giving 40 minutes measures the gap from the missed 09:25 bus to the 10:05 bus, counting from the bus Chloe missed instead of from the time she arrived. Giving 5 minutes uses the missed 09:25 bus, working out 09:30 − 09:25 the wrong way round. Giving 45 minutes comes from treating the times as decimals — 10.05 − 9.30 = 0.75, then × 60 = 45 — instead of working in hours and minutes, where 05 minutes past the hour is not 0.05 of an hour.
- (d) 30 — Method: on a dual bar chart each bar is a separate frequency, so a total for the whole class is found by combining the two frequencies the bars show. Working: the bars show 14 boys and 16 girls, and 14 + 16 = 30. Answer: 30 pupils, a count of pupils in the class. The distractors: 32 comes from doubling the taller bar, 16 + 16, as though the two bars were equal; 28 comes from doubling the shorter bar, 14 + 14, in the same way; 2 comes from finding the difference between the two bars, 16 − 14, which answers how many more girls there are rather than how many pupils there are.
- (a) 23 minutes — Method: the equation of a line of best fit converts a value of x into a predicted value of y, so substitute the known number of pages for x and evaluate. Working: x is the number of pages, so put x = 10 into y = 2x + 3. Multiplication is carried out before addition, so 2 × 10 + 3 = 23. Answer: 23 minutes, and it is a prediction of the trend rather than a promise about any one chapter. The distractors: 20 minutes comes from working out 2 × 10 and stopping there, leaving out the 3 that the line adds; 26 minutes comes from reading the equation as y = 2(x + 3), adding first and then doubling, so 2 × 13 = 26; 13 minutes comes from adding 10 and 3 and never using the gradient at all, which treats the 2 as though it were not there.
- (d) Route 1, as its times vary by 6 minutes rather than 20 — Method: work out an average and a measure of spread for each route, then decide which matters to a commuter who must arrive on time every day. Working: for Route 1, 22 + 23 + 24 + 24 + 25 + 25 + 26 + 26 + 27 + 28 = 250 and 250 ÷ 10 = 25, so the mean is 25 minutes, and the range is 28 − 22 = 6 minutes. For Route 2, 18 + 19 + 20 + 20 + 21 + 22 + 26 + 30 + 36 + 38 = 250 and 250 ÷ 10 = 25, so the mean is also 25 minutes, but the range is 38 − 18 = 20 minutes. The means give no reason to prefer either route; the spreads do, because a commuter who must never be late has to allow for the worst day, which is 28 minutes on Route 1 and 38 minutes on Route 2. Answer: Route 1, as its times vary by 6 minutes rather than 20. The distractors: saying Route 2 has the lower mean assumes that its quicker-looking early times must pull the average down, when both routes total 250 minutes over the ten days; choosing Route 2 for its fastest journey of 18 minutes judges a route by its best day, and the commuter has to survive its worst; saying either route will do uses the equal means and ignores the spread altogether, which is the one thing that separates the two routes.
- (c) 16 kg — Method: sort the seven weights before finding the middle value. Working: in order, the weights are 10, 12, 14, 16, 18, 20 and 50 kg. There are 7 values, so the median is the 4th one: 16 kg. Reading off the 4th weight in the order the vet recorded them, 10 kg, skips the sorting step and is not the median. Working out the mean, 140 ÷ 7 = 20 kg, finds a different average altogether. Working out the range, 50 − 10 = 40 kg, finds the spread, not the middle value. Always sort your data first — the median lives in the ordered list, not the collection order.
- (c) No, 90 cm is far outside the heights on the graph — Method: a line of best fit describes the trend only across the stretch of data it was drawn through; predicting beyond that stretch is extrapolation, and nothing in the data supports it. Working: the heights used to draw this line run from 150 cm to 180 cm, all of them Year 10 pupils, while 90 cm is 60 cm below the shortest of them and belongs to a two-year-old child, whose build follows no trend the graph has measured. Substituting anyway gives 0.9 × 90 − 85 = −4, a mass of −4 kg, which cannot exist. Answer: no, because 90 cm is far outside the heights on the graph. The distractors: saying a line of best fit cannot be used to predict at all throws away its main purpose, since a prediction made between the plotted values is perfectly sound; saying the line passes through all 20 points misdescribes a line of best fit, which is drawn to follow the trend of the points and will normally pass through few of them; saying the equation works for any value put into it treats an equation fitted to Year 10 heights as a law of nature, and the mass of −4 kg shows what that assumption produces.
- (d) 1/2 — Method: a percentage is turned into a fraction by writing it over 100 and then cancelling the fraction down to its simplest form. Working: the reading sector is 50% of the pie chart, so as a fraction it is 50/100; dividing the numerator and the denominator by 50 gives 1/2. Answer: 1/2 of the pie chart — a fraction of the chart, not a number of pupils. The distractors: 1/4 comes from reading the share of the sport sector, 25%, instead of the share of the reading sector; 1/3 comes from counting the three sectors and assuming that three sectors must each take a third of the chart, which is only true when the sectors are equal; 3/4 comes from adding the reading and sport sectors together, 50% + 25% = 75%, instead of taking the reading sector on its own.
- (c) Thursday — Method: on a bar chart the tallest bar belongs to the greatest value, so compare the five temperatures and then read off the day that the greatest one belongs to. Working: the temperatures are 20 °C, 22 °C, 18 °C, 25 °C and 23 °C; in order of size these are 18, 20, 22, 23 and 25, so the greatest temperature is 25 °C, and the day recorded with 25 °C is Thursday. Answer: Thursday — the answer is a day, not a temperature. The distractors: Friday comes from stopping at the final temperature listed instead of comparing all five; Wednesday comes from picking out the shortest bar, 18 °C, and so answering for the lowest temperature rather than the highest; Monday comes from writing down the first day in the chart without comparing any of the temperatures at all.
- (c) A line of best fit — Method: the straight line drawn on a scatter graph is named from the job it does — it is chosen so that it follows the whole set of points as closely as possible. Working: the line passes through the middle of the points, with roughly as many points above it as below it, and it need not pass through any of the plotted points at all; the name given to the straight line chosen in that way is a line of best fit. Answer: a line of best fit. The distractors: a line of symmetry comes from confusing a trend with symmetry, which is a property of a shape rather than of a set of data; a horizontal line through the mean comes from thinking the trend is shown by an average, when a horizontal line would say that the vertical quantity does not change and so show no correlation; a line joining the first and last points comes from thinking the line must join the two extreme points, which lets two points decide a trend that all of the points should share in.
- (d) The modal class, as the class with most pupils is shown — Method: a grouped frequency table records how many values fall into each class, but not the values themselves, so any average that needs the individual times can only be estimated from it. Working: the four frequencies are 8, 12, 6 and 4, and 8 + 12 + 6 + 4 = 30, so every pupil is counted. The largest frequency is 12, which belongs to the class 10 < t ≤ 20, and that class can be written down exactly, because finding it needs nothing but the counts the table already gives. Answer: the modal class, as the class with most pupils is shown. The distractors: the mean is said to use all 30 times, but the table does not hold them; the usual method replaces each class by its midpoint, 5, 15, 25 and 35, which gives an estimate of the mean and not its true value; the median is said to be shown, but the table locates only the class holding the 15th and 16th times, which is 10 < t ≤ 20, without saying what either time was; the range is said to be shown, but 0 and 40 are the boundaries of the first and last classes, not the fastest and slowest times actually recorded.
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