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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Statistics worksheet — GCSE Foundation
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- 1.Two classes at a school in Coventry sit the same maths test, out of 20 marks. Class A has a mean mark of 14 and a range of 6. Class B has a mean mark of 14 and a range of 14. Write a sentence comparing the two classes, using the mean and the range.
- 2.A scatter graph shows the number of pages, x, in a chapter and the time, y minutes, a pupil took to read it, for 15 chapters. The line of best fit has equation y = 2x + 3. Use the line of best fit to estimate the time taken to read a chapter of 10 pages.y = 2x + 3
- 3.A scatter graph has 50 points. Most of them lie close to a rising line of best fit, but two of them lie a long way from that line. Write down how those two points should be treated.
- 4.A factory makes 4,000 light bulbs a day. In a random sample of 80 of one day's bulbs, 3 were faulty. Work out an estimate for the number of faulty bulbs the factory makes in a day.
- 5.A scatter graph plots the shoe size and the spelling test score of 25 pupils. The points are scattered with no pattern across the graph. Write down the type of correlation shown.
- 6.Work out the median of these five numbers: 2, 4, 7, 12, 26
- 7.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.
- 8.Isla wants to find out the favourite television programme of the pupils at her school. She asks only her own close group of friends. Write down what is wrong with her sample.
- 9.A teacher records the number of pets owned by each of 25 pupils in a class; each pupil owns 0, 1, 2, 3 or 4 pets. The teacher wants to show how many pupils own each number of pets. Write down the most suitable type of chart for this data, and give a reason for your answer.
- 10.A café owner in Brighton records the midday temperature, x °C, and the number of hot chocolates sold, y, on 12 days. The temperatures recorded run from 4 °C to 18 °C, and the line of best fit is y = −3x + 74. The forecast for tomorrow gives a midday temperature of 12 °C. Work out the number the line of best fit predicts, and write down how much confidence the owner can have in it.y = -3x + 74
- 11.A pie chart of the favourite hobby of a group of pupils has three sectors: reading 50%, sport 25% and music 25%. Write down, as a fraction in its simplest form, the share of the pie chart taken up by the reading sector.
- 12.A café sold 10 sandwiches on each of four days: 10, 10, 10, 10. Work out the range of the numbers sold.
- 13.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.
- 14.The weekly wages of the five people who work at a small garage in Norwich are £420, £440, £460, £480 and £1,500. Write down which average better describes a typical wage at this garage, and give a reason for your answer.
- 15.A random sample of 50 pupils at a school were asked whether they prefer sport to music. 30 of the 50 pupils said they prefer sport. The school has 500 pupils altogether. Work out an estimate for the number of the 500 pupils who prefer sport.
Answer key
- (a) Equal means; Class A is more consistent, smaller range. — Method: when two data sets share a measure of location, compare a measure of spread to say more about consistency. Working: both classes have the same mean mark, 14, so on average they performed equally well. Class A has the smaller range, 6, so its marks are more tightly grouped around 14 than Class B's marks, which vary by as much as 14. So Class A's marks were more consistent, even though neither class did better on average. Saying Class B did better because it has the bigger range confuses a wide spread with a high score — a big range describes variability, not performance. Saying Class A did better because it has the smaller range makes the same mistake in the other direction: the two classes are tied on the mean, so neither one 'did better'. Saying the classes cannot be compared because their means are equal misses the whole point of also comparing the range. Always compare both an average AND a spread before describing two data sets — either one alone tells only half the story.
- (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.
- (b) Treat them as outliers and check them before deciding — Method: a point lying a long way from the pattern the rest of the data make is called an outlier, and an outlier is investigated before anything is done with it, because it may be an error in the data or it may be a genuine but unusual case. Working: 48 of the 50 points lie close to the rising line of best fit, so the trend is set by those 48; the two remaining points do not follow it, so they are identified as outliers and checked — a mistake in measuring or recording would be corrected, while a genuine reading would be kept and reported. Answer: treat them as outliers and check them before deciding what to do with them. The distractors: deleting them at once assumes that every point far from the line must be an error, which throws away real data; moving the line so that it passes through them assumes a line of best fit must touch particular points, when it is drawn to follow all 50; taking them as proof that there is no correlation lets two points overturn the pattern that the other 48 agree on.
- (a) 150 bulbs — Method: assume the proportion faulty in a random sample is the proportion faulty in the whole day's output, and scale the sample up to the population. Working: the sample of 80 has to be scaled up to 4,000 bulbs, and 4,000 ÷ 80 = 50, so the day's output is 50 sample-sized batches. Each batch is expected to contain the same 3 faulty bulbs, so the estimate is 3 × 50 = 150. Answer: 150 bulbs, and it is an estimate, because another sample of 80 would probably contain a different number of faulty bulbs. The distractors: 50 bulbs is the scale factor 4,000 ÷ 80 written down as though it were the answer, so it reports how many batches there are rather than how many faulty bulbs; 120 bulbs comes from reading 3 out of 80 as 3%, then taking 0.03 × 4,000 = 120, but 3 out of 80 is 3.75%; 240 bulbs comes from 3 × 80 = 240, multiplying the faulty bulbs by the size of the sample instead of by the scale factor, which uses the 80 twice and the 4,000 not at all.
- (a) No correlation — Shoe size has no real relationship with spelling ability, and the points here are scattered with no rising or falling trend, so this is no correlation. A positive correlation would show the points rising together, and a negative correlation would show them falling as one increases; neither pattern is present here. Strong correlation is not correct either, since strength only applies once a positive or negative trend exists, and there isn't one.
- (c) 7 — Method: the median is the middle value when the data are written in order of size, and with an odd number of values there is exactly one middle value. Working: the numbers are already in order, 2, 4, 7, 12, 26, and there are 5 of them, so the middle position is the third and the value sitting there is 7. Answer: 7, with two values below it and two above it. The distractors: 10.2 comes from working out the mean, 51 ÷ 5, instead of the median; 14 comes from taking the value halfway between the smallest and the largest, (2 + 26) ÷ 2; 24 comes from working out the range, 26 − 2, which measures spread rather than centre.
- (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.
- (c) It is not representative, as she picked her own friends — Method: judge a sample by asking whether the pupils in it were chosen in a way that gives the whole school a fair chance of being heard. Working: Isla's friends are a group she formed herself, and friends tend to share tastes, so their favourite programme is likely to match hers rather than the school's, and pupils in other year groups and other friendship groups had no chance at all of being asked; the fault lies in how the pupils were selected, not in how many of them there were. Answer: it is not representative, as she picked her own friends. The distractors: the reply blaming the size claims the pupils were picked at random, which is false here, and it is the common mistake of thinking a biased sample can be cured by making it bigger; the reply calling the sample too large is false in the other direction, as a survey is never spoilt by collecting more replies; the reply that the sample is fine treats attending the school as enough, which would make any group of pupils in the building a fair sample.
- (a) A vertical line chart (discrete numerical data) — The number of pets is discrete numerical data — whole-number values such as 0, 1, 2, 3 or 4 — recorded for one variable, so a vertical line chart is the chart specified for this kind of data. A bar chart is used for categorical data, such as favourite colour, not numerical values counted like this. A pie chart shows proportions of a whole and does not show the frequency of each separate value. A scatter graph compares two different variables against each other, and only one variable, the number of pets, is recorded here.
- (d) 38, and fairly confident, as 12 °C is inside the range — Method: substitute the forecast temperature into the equation of the line of best fit, then judge the prediction by where that temperature sits among the data the line was drawn from. Working: putting x = 12 into y = −3x + 74 gives −3 × 12 + 74 = 38, so the line predicts 38 hot chocolates. The recorded temperatures run from 4 °C to 18 °C, and 12 °C lies inside that interval, so this is interpolation, the safer kind of prediction. Answer: 38, and fairly confident, as 12 °C is inside the range; the owner should still expect the true figure to differ a little, since the points only lie near the line and not on it. The distractors: being completely certain treats a line of best fit as a rule that fixes each day's sales, when it describes a trend that individual days depart from; saying 12 °C is outside the range misreads the interval 4 °C to 18 °C, and the wrong warning would be attached to a sound prediction; 110 comes from −3 × 12 being taken as +36, giving 36 + 74 = 110, which loses the negative gradient and so predicts that a warm day sells more hot chocolate than a cold one.
- (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.
- (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.
- (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.
- (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.
- (d) 300 pupils — Method: an estimate for a whole population is made by finding the proportion in the sample and applying that same proportion to the population. Working: in the sample 30 of the 50 pupils prefer sport, a proportion of 30 ÷ 50 = 0.6, and applying that proportion to the school gives 0.6 × 500 = 300 pupils. Answer: 300 pupils, and it is only an estimate, because a different random sample of 50 would give a slightly different figure. The distractors: 200 pupils comes from scaling up the 20 pupils in the sample who did not prefer sport, 20 × 10, which answers the opposite question; 150 pupils comes from reading 30 out of 50 as 30% and taking 30% of 500; 60 pupils comes from working out the proportion correctly as 60% and then writing the 60 down as a number of pupils instead of applying it to the 500.
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