Printable · GCSE Higher · ages 14-16
Statistics worksheet — GCSE Higher
Fifteen questions across the statistics statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Answer key: Statistics worksheet — GCSE Higher
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- (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.
- (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.
- (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) On average a plant grew 1.5 cm taller for each extra day — Method: in the equation of a line, the number multiplying x is the gradient, and a gradient states the change in y produced by an increase of 1 in x, read in the units of the two axes. Working: here x is measured in days and y in centimetres, so the gradient 1.5 carries the units centimetres per day. Testing it on the line, 5 days gives 1.5 × 5 + 4 = 11.5 cm and 6 days gives 1.5 × 6 + 4 = 13 cm, a rise of 1.5 cm for the one extra day. Answer: on average a plant grew 1.5 cm taller for each extra day of watering. The distractors: 1.5 cm as the height before any watering is the value of y when x is 0, which is the other number in the equation, 4 cm, so this swaps the gradient and the intercept; 1.5 cm as the gap between the tallest and the shortest plant reads the gradient as a range, when a range is a difference between two of the 16 plants and a gradient is a rate; 1.5 days for each extra centimetre inverts the rate, dividing days by centimetres instead of centimetres by days, and the line gives 1 cm of growth in two thirds of a day.
- (a) 22 — Method: the two subject totals overlap, because every pupil who passed both subjects has been counted once in the maths total and once again in the science total; adding the totals therefore counts those pupils twice, and the overlap has to be taken off once. Working: 18 + 12 = 30, and the 8 pupils who passed both have been counted twice in that 30, so the number who passed at least one subject is 30 − 8 = 22. Answer: 22 pupils, a count of pupils, and it is less than the 30 in the class, which leaves 8 pupils who passed neither. The distractors: 30 comes from adding the two subject totals and never removing the overlap, so it counts the 8 pupils twice; 14 comes from taking the 8 away twice, 18 + 12 − 8 − 8, removing an overlap that was only counted twice once too often; 18 comes from writing down the larger of the two subject totals on its own, which leaves out every pupil who passed science but not maths.
- (c) 75 — Method: the number in a class is the area of its bar, frequency density × class width, so work out the frequency of each class that lies at or above 10 minutes and add them. Working: the class 10 ≤ t < 25 is 15 minutes wide with a frequency density of 3.2, giving 3.2 × 15 = 48 members; the class 25 ≤ t < 55 is 30 minutes wide with a frequency density of 0.9, giving 0.9 × 30 = 27 members; the total charged is 48 + 27 = 75. Answer: 75 members pay the extra charge. The distractors: 4.1 comes from adding the two frequency densities, 3.2 + 0.9, as though each height were a count; 93 comes from including the class 0 ≤ t < 10 as well, 1.8 × 10 = 18 added to 48 and 27, which charges every member; 27 comes from using only the class 25 ≤ t < 55 and forgetting that 10 ≤ t < 25 is also at or above 10 minutes.
- (b) 6 — Method: frequency density = frequency ÷ class width. Working: the class 12 ≤ h < 18 has width 18 − 12 = 6, so frequency density = 36 ÷ 6 = 6. Answer: the frequency density is 6 seedlings per cm. Watch which numbers you use: taking the lower bound, 12, as the width instead of 18 − 12 = 6 gives 36 ÷ 12 = 3; dividing the total number of seedlings, 90, rather than this class's frequency, 36, by the width gives 90 ÷ 6 = 15, a density that belongs to no single class; and multiplying instead of dividing gives 36 × 6 = 216, far too large a density for so narrow a class.
- (d) 210 — 350 − 150 = 200. 200 ÷ 10 = 20, so the gradient is 20. Using the point (5, 150): 20 × 5 = 100, so 150 − 100 = 50 is the intercept, giving the line y = 20x + 50. At x = 8: 20 × 8 = 160, and 160 + 50 = 210, so the estimated number of visitors is 210. Choosing 160 stops after 20 × 8 = 160 and forgets to add the intercept of 50. Choosing 250 comes from averaging the two given y-values: 150 + 350 = 500, and 500 ÷ 2 = 250, instead of using the line's equation. Choosing 240 assumes the visitors are directly proportional to the hours of sunshine using the first point, 150 × 8 ÷ 5 = 240, which ignores that the line does not pass through the origin.
- (b) 60 — Method: for n ordered values, the upper quartile sits at position 3(n + 1) ÷ 4, counting from the smallest. Working: n + 1 = 11 + 1 = 12; 3 × 12 = 36 and 36 ÷ 4 = 9, so the upper quartile is the 9th value in the list 40, 42, 45, 47, 50, 52, 55, 58, 60, 63, 65, which is 60. Answer: the upper quartile is 60 seconds. Watch which quartile you find: counting to the 6th value gives the median, 52, not the upper quartile; using 3 × 11 = 33 and 33 ÷ 4 = 8.25 without adding 1 to n first, then rounding down, reaches the 8th value, 58, not the 9th; and counting to the 3rd value uses the lower quartile's position, 45, the wrong end of the list.
- (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.
- (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) A histogram, with frequency density up the vertical axis — Method: decide which diagram makes area stand for frequency, which is the property the question asks for. Working: on a histogram the vertical axis is frequency density, so the area of a bar is frequency density × class width, and that product is the frequency; this is exactly what is wanted, and it is what allows classes of unequal width to be shown fairly. Answer: a histogram, with frequency density up the vertical axis. The distractors: a bar chart plots frequency as the height, so with unequal widths a wide class would cover far more area than a narrow class holding the same number of batteries, and area would measure nothing; a cumulative frequency diagram plots running totals against upper class boundaries, so a point on it gives how many lie below a value rather than how many lie in a class; a pie chart shows each class as a share of the whole 300 and loses the class widths entirely, so no area on it is tied to a scale of hours.
- (d) 19 kg — Method: estimate the mean of grouped data by multiplying each class's midpoint by its frequency, adding the four totals, then dividing by the total frequency. Working: the midpoints are 5, 15, 25 and 35 kg. The weighted totals are 11 × 5 = 55, 5 × 15 = 75, 5 × 25 = 125 and 9 × 35 = 315, which add to 570. Dividing by the 30 dogs gives an estimate of 570 ÷ 30 = 19 kg. Giving 5 kg reads off the midpoint of the modal class, 0 < m ≤ 10, the class with the most dogs — but the class with the most dogs is not where the mean falls, and neither is a substitute for actually calculating it. Giving 20 kg averages the four midpoints, (5 + 15 + 25 + 35) ÷ 4, treating every class as equally likely and ignoring that far more dogs are in the lightest and heaviest classes than in the middle two. Giving 570 kg stops after finding the correct weighted total and forgets the final division by the 30 dogs. Always weight each midpoint by its own frequency, and always finish by dividing by the total frequency, not the number of classes.
- (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.
- (b) 120, unreliable — x = 25 is outside 1 to 12 — The line of best fit is y = 4x + 20. 4 × 25 = 100, and 100 + 20 = 120, so the estimate is y = 120. But x = 25 lies far outside the plotted range of 1 to 12 years, so this is an extrapolation, and the estimate is not reliable. Reaching 100 instead of 120 comes from 4 × 25 = 100 with the intercept of 20 left out — still correctly flagged as unreliable, but the wrong value. Calling the estimate reliable simply because it was calculated correctly, giving 120, wrongly assumes that a correct calculation is automatically trustworthy, ignoring that x = 25 lies far beyond the data actually collected. Reaching 68, from 4 × 12 = 48 and 48 + 20 = 68, substitutes x = 12, the top of the plotted range, instead of the assistant's actual x = 25, and wrongly calls that reliable because 12 lies inside the range.
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