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.
Answer key: Statistics worksheet — GCSE Foundation
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- (d) £15.50 — 2 × 6 = 12, and 12 + 3.50 = 15.50, so the estimated fare is £15.50. Choosing £12.00 stops after 2 × 6 = 12 and forgets to add the £3.50. Choosing £19.00 adds the distance and the constant first and then multiplies: 6 + 3.50 = 9.50, and 9.50 × 2 = 19.00, applying the ×2 to the whole sum instead of only to the distance. Choosing £13.00 multiplies only the constant term by 2 instead of the distance: 2 × 3.50 = 7, and 7 + 6 = 13.00.
- (a) 3/10 — There are 60 − 9 = 51 right-handed pupils. Of the 24 pupils who prefer Art, 6 are left-handed, so 24 − 6 = 18 are right-handed and prefer Art. The probability that a randomly chosen pupil is right-handed and prefers Art is 18/60, which simplifies to 3/10. Giving 2/5 is 24/60 simplified — the probability of preferring Art, ignoring the right-handed condition entirely. Giving 17/20 is 51/60 simplified — the probability of being right-handed, ignoring the Art condition entirely. Giving 1/10 is 6/60 simplified — the probability of being left-handed and preferring Art, the wrong hand condition.
- (a) 144° — Method: a pie chart shares the 360° at its centre between the categories in proportion to their frequencies, so the angle of a sector is that category's fraction of the total multiplied by 360°. Working: the sector stands for 40 pupils out of 100, which is the fraction 40/100; one pupil is worth 360 ÷ 100 = 3.6°, so 40 pupils are worth 40 × 3.6° = 144°. Answer: 144°, and the answer is an angle in degrees, not a number of pupils. The distractors: 40° comes from sharing out 100 instead of 360, so the percentage is written straight down as a number of degrees; 216° comes from working out the angle for the other 60 pupils, 60 × 3.6°, which is the rest of the pie chart; 180° comes from assuming that the tallest line must stand for half of the pupils and so take half of the chart.
- (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.
- (c) Strong negative correlation — Method: correlation is described by two things — the direction the points take as the graph is read from left to right, and how closely the points lie to a single straight line. Working: reading the pairs in order of age, the ages rise 14, 18, 23, 27, 31, 36, 42, 49 while the scores fall 92, 88, 85, 80, 78, 74, 70, 65; the score falls at every single step, with no reversal anywhere, so the points fall from left to right and lie close to a straight line. Answer: strong negative correlation — negative for the falling direction, strong because every point follows the pattern. The distractors: strong positive correlation comes from noticing a clear pattern and calling any clear pattern positive, without checking the direction; weak negative correlation comes from reading the direction correctly but judging points that do not lie exactly on a straight line to be only loosely related, when these eight fall without a single exception; no correlation comes from reading a falling trend as though it showed no relationship at all, when a falling trend is itself a relationship.
- (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.
- (b) 74 marks — Method: the two groups are different sizes, so their means cannot simply be averaged — rebuild each group's total mark, add the totals and divide by all 50 pupils. Working: Group A scored 20 × 80 = 1600 marks and Group B scored 30 × 70 = 2100 marks, giving 1600 + 2100 = 3700 marks altogether, so the overall mean is 3700 ÷ 50 = 74 marks. Answer: 74 marks, which sits nearer to 70 than to 80 because the larger group scored 70. The distractors: 75 marks comes from averaging the two group means, (80 + 70) ÷ 2, as though the groups were the same size; 76 marks comes from attaching each mean to the other group's size, (20 × 70 + 30 × 80) ÷ 50; 150 marks comes from adding the two means together and never dividing at all.
- (b) 29 — There are 40 − 24 = 16 males, and 15 of them prefer cardio, so 16 − 15 = 1 male prefers weights. There are 24 females, and 10 prefer weights, so 24 − 10 = 14 females prefer cardio. Altogether, 15 + 14 = 29 people prefer cardio. Choosing 15 only counts the males who prefer cardio and forgets the females. Choosing 11 adds the two weights figures, 1 + 10 = 11, instead of the two cardio figures. Choosing 30 comes from 40 − 10, subtracting only the number of females who prefer weights from the grand total, rather than finding both cardio sub-totals separately.
- (d) 45 — Method: convert the angle into a fraction of the full circle, 360°, then apply that fraction to the total number of shoppers. Working: the card sector is 90° out of 360°, a fraction of 90 ÷ 360 = 0.25. Applying that fraction to the 180 shoppers gives 0.25 × 180 = 45 shoppers. Giving 90 states the angle itself, not a number of shoppers — the angle first has to be converted into a fraction. Using the remaining angle, 360 − 90 = 270°, and scaling that, 270 ÷ 360 × 180 = 135, finds the number who did NOT pay by card, not the number who did. Dividing 360 by 90, 360 ÷ 90 = 4, finds how many equal 90° sectors fit in the circle, a fact about the pie chart's shape, not about the shoppers at all. Always convert the angle to a fraction of 360° first, and apply that same fraction to the total number of people.
- (a) 1.24 — Method: for data given as a frequency table, the mean is Σfx ÷ Σf — multiply each value by its frequency, add the results, then divide by the total frequency. Working: 0 × 6 = 0. 1 × 10 = 10. 2 × 6 = 12. 3 × 3 = 9. So Σfx = 0 + 10 + 12 + 9 = 31. The total frequency is Σf = 6 + 10 + 6 + 3 = 25. Mean = 31 ÷ 25 = 1.24 siblings. Averaging the frequency column itself, (6 + 10 + 6 + 3) ÷ 4 = 6.25, mixes up the frequencies with the values they belong to. Writing down 1, the number of siblings with the highest frequency, gives the mode, not the mean. Writing down 31 stops after finding Σfx and forgets to divide by the total frequency, 25. Always divide Σfx by Σf — never stop at the top of the fraction.
- (a) Minutes a candle has burned and length remaining — As a candle burns for longer, less of it remains, so these two variables move in opposite directions as one increases — that is negative correlation. A pupil's shoe size generally increases as they get older, so age and shoe size show positive correlation, not negative, since both rise together. A football team's shirt colour is not a numerical quantity linked to how many matches it wins, so shirt colour and number of wins show no correlation at all. The number of letters in a pupil's name has no real connection to their ability in maths, so that pair also shows no correlation.
- (d) 25 — In order, the nine ages are 14, 18, 20, 23, 25, 29, 31, 36 and 42, and with 9 values the median is the 5th one, which is 25. Choosing 23 takes the 4th value instead of the 5th. Choosing 29 takes the 6th value instead of the 5th. Choosing 26 comes from averaging the 4th and 6th values, 23 + 29 = 52, and 52 ÷ 2 = 26, a method that is only needed when there is an even number of values.
- (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.
- (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.
- (a) 12 mm — Method: the range is the highest value minus the lowest value. Working: the highest rainfall is 15 mm and the lowest is 3 mm, so the range is 15 − 3 = 12 mm. Giving 15 mm alone states the highest value, not the range. Giving 3 mm alone states the lowest value, not the range. Sorting the six values, 3, 5, 9, 11, 12 and 15, and averaging the middle two, (9 + 11) ÷ 2 = 10 mm, finds the median, a completely different statistic. The range always needs BOTH the highest and the lowest value — never just one of them.
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