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.A company found that the higher the price it charged for a product, the lower the satisfaction score its customers gave. The price and the satisfaction score are plotted on a scatter graph. Describe the line of best fit that would be drawn on that graph.
- 2.A factory made 3,000 phone cases last week. Shift A checked a random sample of 100 cases and found that 34 were scratched. Shift B checked a different random sample of 50 cases and found that 21 were scratched. Using the COMBINED results from both shifts, work out an estimate for the number of scratched cases made last week.
- 3.A company makes 50,000 light bulbs a day and wants to check how long they last before they fail. Testing a bulb to find out how long it lasts destroys it. Give a reason why the company should test a sample of bulbs rather than every bulb it makes.
- 4.A charity shop in Leicester made £2,400 in one month. A pie chart shows how this was spent: repairs took up an angle of 90°, wages took up an angle of 150°, and the rest was other costs. Work out how much was spent on wages.
- 5.A sports centre in Newcastle asks 60 members which activity they prefer: swimming 15 members, football 25 members, badminton 12 members and other activities 8 members. Work out the angle at the centre of the pie chart sector that represents badminton.
- 6.A study found that people who drink more coffee tend to concentrate better at work. A coffee company says that this shows that drinking coffee improves concentration. Give the reason why this conclusion cannot be drawn.
- 7.In a scatter graph of the age, in years, and the wingspan, in cm, of 20 birds of the same species, all the points lie close to a rising line of best fit except one, which lies a long way below the line. That bird was later found to have a damaged wing. Give a reason why this point should not be used when drawing the line of best fit.
- 8.The numbers 2, 4, 6, 8, 10 and 12 have a mean of 7 and a median of 7. The value 100 is now added to the list. Which measure is changed more by adding 100, and why?
- 9.A scatter graph of the number of hours, x, that pupils revised against their test score, y, has the line of best fit y = 2.5x + 15. Amelia wants a score of at least 80. Work out the least whole number of hours of revision the line of best fit suggests she needs.y = 2.5x + 15
- 10.A scatter graph plots the number of years, x, that 20 employees have worked at a company against their salary, y. All the plotted points lie between x = 1 and x = 15. Write down the word used to describe an estimate for y made using a value of x that lies between 1 and 15.
- 11.A council in Leeds wants to know what local people think about letting shops stay open later in the evening. It rings landline telephone numbers between 10 am and 2 pm on a Tuesday. Write down which group is most likely to be under-represented in the sample, and give a reason for your answer.
- 12.The table shows the times, t minutes, that 30 pupils took to walk to school. 0 < t ≤ 10: 8 pupils. 10 < t ≤ 20: 12 pupils. 20 < t ≤ 30: 6 pupils. 30 < t ≤ 40: 4 pupils. Write down which average can be given exactly from this table, and give a reason for your answer.
- 13.On a scatter graph the horizontal axis shows height in centimetres and the vertical axis shows mass in kilograms. One point is plotted at (170, 65). Write down the height and the mass of that person.
- 14.A bar chart is drawn for a set of categorical data. Write down what the height of each bar represents.
- 15.A recruitment agency in Manchester compares the weekly pay, in pounds, of seven employees at two branches. Branch A: £480, £495, £500, £505, £510, £515, £1,200. Branch B: £480, £490, £500, £510, £520, £530, £540. An advert for the agency claims 'Branch A pays more on average.' Decide whether this claim is fairly supported, using an appropriate average, and choose the correct conclusion.
Answer key
- (a) A straight line sloping down from left to right — Method: a line of best fit is a straight line drawn to follow the trend of the points, so its slope is decided by the direction of the relationship between the two quantities. Working: as the price rises, the satisfaction score falls, so the points start high on the left of the graph and finish low on the right; the straight line that follows them therefore slopes downwards as the graph is read from left to right, which is the line of a negative correlation. Answer: a straight line sloping down from left to right. The distractors: a line sloping up comes from reading a falling relationship as a rising one; a horizontal line comes from expecting no correlation, since a horizontal line says the satisfaction score does not change as the price changes; a curve passing through every point comes from thinking a line of best fit has to touch all of the plotted points, when it is a single straight line drawn through the middle of them.
- (d) 1100 — Method: to combine two samples of different sizes, add the faulty counts together and add the sample sizes together before scaling up, rather than treating the two samples separately. Working: the combined sample found 34 + 21 = 55 scratched cases out of 100 + 50 = 150 cases checked, a proportion of 55 ÷ 150. Applying that proportion to the week's production of 3,000 gives an estimate of 55 ÷ 150 × 3000 = 1100 scratched cases. Averaging the two shifts' proportions instead of combining their totals, (34 ÷ 100 + 21 ÷ 50) ÷ 2 = 0.38, gives 0.38 × 3000 = 1140 — this treats the two samples as equally weighted even though Shift A checked twice as many cases as Shift B. Using only Shift A's sample, 34 ÷ 100 × 3000 = 1020, ignores Shift B's cases completely. Using only Shift B's sample, 21 ÷ 50 × 3000 = 1260, ignores Shift A's cases completely. When two samples are different sizes, combine their totals before finding the proportion — do not average the two proportions, and do not use only one shift's sample.
- (d) Testing destroys bulbs, so testing all leaves none to sell. — Method: testing every item in a population instead of a sample is a census — sensible only when testing does not use up or destroy what is being tested. Working: here, testing a bulb to find its lifespan destroys it, so testing all 50,000 bulbs would leave nothing left to sell — a sample lets the company estimate the typical lifespan without destroying its whole stock. Extra electricity used in testing is not the real reason a census is avoided here — it is the destruction of the product that matters. Saying a sample is always more accurate than a full census is the wrong way round: a census, if it could be carried out, gives the exact figure for the whole population — it is testing being destructive, not a lack of accuracy, that rules it out here. There is no law against testing every item a company makes — nothing in the question suggests that. When testing destroys the item being tested, sampling is necessary, not just convenient.
- (b) £1,000 — Wages take up 150° out of 360°, so the amount spent on wages is 150 ÷ 360 × 2400 = £1,000. Choosing £600 uses the repairs angle, 90°, instead of the wages angle: 90 ÷ 360 × 2400 = 600. Choosing £3,600 treats the angle in degrees as if it were a percentage, 150 ÷ 100 × 2400 = 3600, instead of dividing by 360°. Choosing £800 uses the angle for the 'other costs' sector, 360 − 90 − 150 = 120°, instead of the wages sector: 120 ÷ 360 × 2400 = 800.
- (c) 72° — The angle is 12 ÷ 60 × 360 = 72°. Choosing 150° divides football's frequency of 25 instead of badminton's 12: 25 ÷ 60 × 360 = 150. Choosing 20° finds badminton as a percentage of the members, 12 ÷ 60 × 100 = 20, rather than an angle in degrees. Choosing 90° uses 48, the total of the other three activities, as the total instead of the full 60 members: 12 ÷ 48 × 360 = 90.
- (c) The data show a link only; a third factor may affect both — Method: a study of this kind measures two quantities and reports how they change together; deciding that one of them produces the other is a further claim, and it needs evidence that the measurements alone cannot give. Working: the study shows that more coffee goes with better concentration, which is a positive correlation; but a third factor that was never measured, such as how motivated someone is, could raise both the coffee drinking and the concentration, and the concentration could equally be what leads to the extra coffee. Answer: the data show a link only, because a third factor may be affecting both quantities, so no claim about cause can be made. The distractors: calling the conclusion safe because the correlation is positive treats the direction of a correlation as proof of cause, which no direction can give; calling it wrong because the correlation is negative misreads the direction of the relationship, since the study reports both quantities rising together; saying the two quantities are not linked denies the correlation the study actually found, when what fails is only the claim about cause.
- (d) An outlier from the damaged wing, not the trend. — That bird's point lies a long way from the rising trend followed by every other bird, and its low wingspan is explained by the damaged wing rather than by its age — it is an outlier caused by an unusual factor, not part of the general relationship between age and wingspan, so it should not be used when drawing the line of best fit. Saying every point must be used ignores that an outlier caused by a separate, identifiable factor can rightly be set aside. Saying it shows no correlation ignores that the other 19 points do show a clear rising trend; one outlier does not remove that. Saying it proves the line is inaccurate confuses one unusual bird with a fault in the line itself, when the line correctly describes the trend followed by the rest of the data.
- (c) The mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3. — Method: work each measure out before the extra value is added and again afterwards, then compare the size of the two changes. Working: before, the six values total 42, so the mean is 42 ÷ 6 = 7, and the middle pair 6 and 8 give a median of (6 + 8) ÷ 2 = 7; after, the seven values total 142, so the mean is 142 ÷ 7 = 20.29 to 2 decimal places, while the median is now the 4th of the seven ordered values, which is 8; the mean has moved by about 13.3 and the median by 1. Answer: the mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3 — this is why the median is often preferred when a data set contains an outlier. The distractors: the reply that the mean rises by 100 adds the extra value to the mean instead of adding it to the total; the reply that the median moves to 12 takes the largest of the original values as the new middle instead of counting to the 4th of the seven values; the reply about even and odd counts quotes a rule that does not exist, since the median moved because a very large value was added, not because the count of values changed.
- (a) 26 — Method: a line of best fit lets one quantity be predicted from the other, so the score is substituted into the equation of the line and the resulting inequality is solved for the number of hours. Working: a score of at least 80 means 2.5x + 15 ≥ 80; taking 15 from both sides gives 2.5x ≥ 65, and dividing both sides by 2.5 gives x ≥ 26, so the least whole number of hours is 26. Checking, 2.5 × 26 + 15 = 80, which does reach the target. Answer: 26 hours — and this is only an estimate, because a line of best fit predicts a trend rather than an individual result, and a prediction made outside the range of hours the pupils actually revised for would be an extrapolation and less reliable still. The distractors: 27 comes from reaching 26 and then rounding up again, although 26 hours already gives a score of exactly 80; 32 comes from 80 ÷ 2.5, which ignores the 15 in the equation of the line; 38 comes from (80 + 15) ÷ 2.5, that is from adding the 15 instead of subtracting it when rearranging.
- (b) Interpolation — The salary is being estimated for a value of x between 1 and 15, which is inside the range of x-values that were actually plotted, so this is interpolation. Extrapolation would apply if the estimate used a value of x below 1 or above 15, outside the plotted range. Correlation describes the relationship between the two variables, not the reliability of an estimate, and causation describes one variable actually causing a change in the other, which is a different idea altogether — neither is the word being asked for here.
- (d) Full-time workers, as most are at work at that time — Method: a sample is biased when the method of contact makes part of the population much less likely to be reached, so test each group against where its members actually are between 10 am and 2 pm on a weekday, and test each stated reason against the facts. Working: those hours are the middle of the working day, so people in full-time employment are at work and not beside a landline telephone, while people who are retired and people who are unemployed are far more likely to be at home and are reached at the usual rate; the method therefore collects far fewer replies from full-time workers than their share of the adult population the council is consulting. Answer: full-time workers, as most are at work at that time. The distractors: the reply naming retired people rests on the false claim that most retired people are at work in the daytime, when in fact a daytime call reaches them more easily than anyone; the reply naming unemployed people rests on the false claim that they are out during the day, when they too are among the easiest people to reach by a daytime call; the reply naming children rests on the false claim that children are at home at 11 am on a Tuesday, when they are at school and so are not reached by the call at all, and school-age children are in any case not the adults whose views the council is collecting.
- (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.
- (c) Height 170 cm, mass 65 kg — Method: a point on a scatter graph is written as a pair of coordinates in which the horizontal value is written first and the vertical value second, so each value is matched to the quantity named on its own axis. Working: in (170, 65) the value 170 is the horizontal coordinate and the horizontal axis shows height in centimetres, so the height is 170 cm; the value 65 is the vertical coordinate and the vertical axis shows mass in kilograms, so the mass is 65 kg. Answer: height 170 cm, mass 65 kg, each with the unit named on its own axis. The distractors: height 65 cm and mass 170 kg come from reading the pair the wrong way round, which would describe an impossible person; height 170 cm and mass 170 kg come from reading the horizontal coordinate for both quantities and never using the second number; height 235 cm and mass 105 kg come from combining the two coordinates, 170 + 65 and 170 − 65, instead of reading them separately.
- (c) The frequency of that category — Method: a bar chart for categorical data has one bar for each category, and the vertical scale on which the bars are measured is a count. Working: a bar drawn twice as tall as another tells you that twice as many items of data fell into its category, so the height measures how many items of data belong to that one category, which is exactly what a frequency is. Answer: the height of each bar is the frequency of that category — a count of items of data. The distractors: the number of different categories comes from reading the vertical scale as though it counted the bars, which is shown along the horizontal axis instead; the total of all the data comes from treating one bar as though it stood for the whole data set rather than for one category; the mean of all the data comes from confusing a bar chart with a measure of average, which no single bar can show.
- (c) No — median £505 at A vs £510 at B. — Branch A's seven wages in order are £480, £495, £500, £505, £510, £515 and £1,200, so the median, the 4th value, is £505. Branch B's in order are £480, £490, £500, £510, £520, £530 and £540, so the median is £510. Since £505 is lower than £510, the median wage is not higher at Branch A, so the claim is not fairly supported. Choosing 'Yes — mean £600.71 at A vs £510 at B' uses the mean: 480 + 495 + 500 + 505 + 510 + 515 + 1200 = 4205, and 4205 ÷ 7 = 600.71, a figure pulled upward by the £1,200 outlier that does not represent a typical wage. Choosing 'Yes — median £515 at A vs £510 at B' miscounts the middle position, taking the 6th wage, £515, instead of the correct 4th value, £505. Choosing 'Yes — highest wage £1,200 at A vs £540 at B' compares the highest wage at each branch rather than a measure of the typical, or average, wage.
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