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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.

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Statistics worksheet — GCSE Higher

Maths · GCSE Higher · Statistics · 15 questions

MathsUKwww.geekhero.co.uk

1.The heights, h cm, of 80 plants are grouped like this: 0 ≤ h < 20, 14 plants; 20 ≤ h < 40, 22 plants; 40 ≤ h < 50, 16 plants; 50 ≤ h < 80, 28 plants. Write down the class interval that contains the lower quartile.
(a)40 ≤ h < 50
(b)50 ≤ h < 80
(c)20 ≤ h < 40
(d)0 ≤ h < 20
2.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.
(a)150 bulbs
(b)240 bulbs
(c)120 bulbs
(d)50 bulbs
3.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.
(a)150 pupils
(b)200 pupils
(c)60 pupils
(d)300 pupils
4.Two classes sat the same maths test, both marked out of 100. Class A had a mean mark of 70 and a range of 30 marks. Class B had a mean mark of 70 and a range of 10 marks. Compare the marks of the two classes.
(a)The means are equal, and Class B's marks are the more consistent because its range is smaller.
(b)Class A's mean mark is higher, and Class A's marks are also the more consistent.
(c)The means are equal, and Class A's marks are the more consistent because its range is larger.
(d)Class B's mean mark is higher, and its smaller range makes its marks the more consistent.
5.The delivery times, in minutes, of 15 parcels are given in order: 5, 8, 10, 12, 14, 15, 17, 19, 21, 23, 25, 27, 29, 31, 34. Work out the interquartile range of these times.
(a)−15
(b)29
(c)15
(d)8
6.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.
(a)Include it — it proves the line is inaccurate.
(b)Include it — every point must be used.
(c)It shows age and wingspan have no correlation.
(d)An outlier from the damaged wing, not the trend.
7.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.
(a)£600
(b)£1,000
(c)£3,600
(d)£800
8.The 120 pupils in Year 11 at a school sat a maths test. Their marks m are grouped into classes of unequal width: 0 ≤ m < 40, 12 pupils; 40 ≤ m < 60, 24 pupils; 60 ≤ m < 70, 36 pupils; 70 ≤ m ≤ 100, 48 pupils. A histogram is drawn for these data. Work out the frequency density of the class 40 ≤ m < 60.
(a)2.4 pupils per mark
(b)24 pupils per mark
(c)0.2 pupils per mark
(d)1.2 pupils per mark
9.The times, t seconds, taken by 142 competitors to complete a lap are grouped like this: 0 ≤ t < 10, 20 competitors; 10 ≤ t < 25, 12 competitors; 25 ≤ t < 45, 50 competitors; 45 ≤ t < 75, 60 competitors. A histogram is drawn. Write down the class whose bar is the tallest.
(a)10 ≤ t < 25
(b)45 ≤ t < 75
(c)0 ≤ t < 10
(d)25 ≤ t < 45
10.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
(a)26
(b)27
(c)38
(d)32
11.Box plots summarise house prices, in thousands of pounds, in two towns. Town A: minimum 120, lower quartile 180, median 220, upper quartile 310, maximum 520. Town B: minimum 95, lower quartile 175, median 235, upper quartile 260, maximum 340. A buyer wants the town where house prices are typically higher but more consistent. Which statement is correct?
(a)Town A — higher median and smaller IQR
(b)Town B — higher median and smaller IQR
(c)Town B — higher median and larger IQR
(d)Town A — higher median and larger IQR
12.The mean of 5 numbers is 8. Work out the total of the 5 numbers.
(a)40
(b)1.6
(c)13
(d)8
13.The mean of the numbers x, 20 and 30 is equal to the mean of the numbers 15 and 25. Work out the value of x.
(a)20
(b)−10
(c)10
(d)70
14.A composite (stacked) bar for a charity bake sale in Durham shows the number of cakes sold, split into three types. The bar has a total height of 80 cakes: 34 were sponge cakes, 26 were chocolate cakes and the rest were fruit cakes. Work out the percentage of the cakes sold that were fruit cakes.
(a)20%
(b)42.5%
(c)32.5%
(d)25%
15.A factory makes a batch of 4,000 circuit boards. It checks a random sample of 50 boards and finds that 4 are faulty. The factory will scrap the whole batch if the estimated number of faulty boards in the batch is more than 250. Should the factory scrap the batch?
(a)No — with an estimate of 80, below the 250 limit.
(b)No — only 4 faulty in the sample, far fewer than 250.
(c)Yes — with an estimate of 320, above the 250 limit.
(d)Yes — with an estimate of 500, above the 250 limit.

Answer key

(c) 20 ≤ h < 40 — Method: with 80 values the lower quartile is the 80 ÷ 4 = 20th value in order, so build a running total until it first reaches 20. Working: the running totals are 14, then 14 + 22 = 36, then 52, then 80; the 20th plant is past 14 but not past 36, so it lies in the second class. Answer: the lower quartile lies in the class 20 ≤ h < 40. The distractors: 0 ≤ h < 20 comes from believing that the bottom quarter of the data must all sit in the first class, when that class holds only 14 of the 80 plants; 40 ≤ h < 50 comes from using the position 80 ÷ 2 = 40 and so locating the median rather than the lower quartile; 50 ≤ h < 80 comes from counting 20 plants down from the tallest instead of up from the shortest, which locates the upper quartile at the 60th plant.
(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.
(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.
(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.
(c) 15 — Method: the interquartile range is the upper quartile take away the lower quartile, IQR = Q3 − Q1. Working: n + 1 = 15 + 1 = 16, so the lower quartile sits at position 16 ÷ 4 = 4, the 4th value in the list, which is 12; the upper quartile sits at position 3 × 4 = 12, the 12th value, which is 27. So 27 − 12 = 15. Answer: the interquartile range is 15 minutes. Watch which values you use and which way round: taking the smallest time away from the largest, 34 − 5 = 29, finds the range, which uses every value between the extremes rather than just the middle half; taking the median away from the upper quartile instead of the lower quartile, 27 − 19 = 8, swaps the median in for the lower quartile; and reversing the subtraction, 12 − 27 = −15, finds the right two values but in the wrong order — an interquartile range is never negative.
(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.
(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) 1.2 pupils per mark — Method: on a histogram whose class intervals are not all the same width the height of a bar is not the frequency but the frequency density, found by dividing the frequency of the class by the width of that class, so that the area of the bar represents the frequency. Working: the class 40 ≤ m < 60 holds 24 pupils, and its width is 60 − 40 = 20 marks, so the frequency density is 24 ÷ 20 = 1.2. Answer: 1.2 pupils per mark. The distractors: 24 pupils per mark comes from plotting the frequency itself as the height, which is only correct when every class has the same width; 2.4 pupils per mark comes from dividing by 10, the width of the narrowest class, instead of by the width of this class; 0.2 pupils per mark comes from dividing by the 120 pupils in the year group, which gives the proportion of pupils in the class and not a frequency density.
(d) 25 ≤ t < 45 — Method: the height of a bar on a histogram is the frequency density, so work out frequency ÷ class width for every class and compare the four heights. Working: 20 ÷ 10 = 2 for the first class; 12 ÷ 15 = 0.8 for the second; 50 ÷ 20 = 2.5 for the third; 60 ÷ 30 = 2 for the fourth. Answer: the largest of 2, 0.8, 2.5 and 2 is 2.5, so the tallest bar is the one for 25 ≤ t < 45. The distractors: 45 ≤ t < 75 comes from picking the class with the greatest frequency, 60, and treating a frequency as a height — but that class is three times as wide, so its 60 competitors are spread thinly; 10 ≤ t < 25 comes from dividing the class width by the frequency, 15 ÷ 12, and picking the largest of those reversed values; 0 ≤ t < 10 comes from assuming the narrowest class must always give the tallest bar, which is only true when the frequencies are equal.
(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) Town B — higher median and smaller IQR — Method: 'higher and more consistent' needs two comparisons — the median for typical price, and the interquartile range for spread, with a smaller interquartile range meaning more consistent. Working: Town B's median, £235,000, is higher than Town A's, £220,000. Town A's interquartile range is 310 − 180 = 130 and Town B's is 260 − 175 = 85, so Town B's interquartile range is the smaller of the two. Answer: Town B has both the higher median and the smaller interquartile range, so it is the town with higher, more consistent prices. Watch which combination of median and interquartile range each statement claims, and for which town: claiming Town A has the higher median and the smaller interquartile range gets both comparisons wrong, since Town B leads on both; claiming Town A has the higher median (still wrong) but the larger interquartile range at least reads the spread correctly, without it rescuing the false median claim; and claiming Town B has the higher median (correct) but the larger interquartile range misreads the spread — Town B's interquartile range is the smaller of the two, not the larger.
(a) 40 — Method: the mean is the total divided by how many values there are, so rearranging gives total = mean × number of values. Working: the mean is 8 and there are 5 numbers, so the total is 8 × 5 = 40. Answer: 40, and checking, 40 ÷ 5 = 8, which is the mean given. The distractors: 13 comes from adding the mean and the count, 8 + 5, instead of multiplying them; 1.6 comes from dividing the mean by the count, 8 ÷ 5, which reverses the relationship; 8 comes from quoting the mean itself as the total, which is only true when there is a single number.
(c) 10 — Method: work out the mean that can be found straight away, then use total = mean × number of values on the group of three to find the missing number. Working: the mean of 15 and 25 is (15 + 25) ÷ 2 = 40 ÷ 2 = 20, so the group of three must also have a mean of 20; three numbers with a mean of 20 have a total of 20 × 3 = 60, and 20 + 30 = 50 of that total is already accounted for, so x = 60 − 50 = 10. Answer: 10, and checking, (10 + 20 + 30) ÷ 3 = 20. The distractors: 20 comes from working out the mean the two groups share and writing that down as x; −10 comes from dividing the group of three by 2 instead of by 3, which gives x + 50 = 40; 70 comes from reading the total 15 + 25 = 40 as the mean of the pair, which sets the target total at 120 and leaves x = 70.
(d) 25% — First find the number of fruit cakes: 80 − 34 − 26 = 20. Then write this as a percentage of the total: 20 ÷ 80 × 100 = 25%. Giving 20% comes from reporting the count of fruit cakes, 20, directly as a percentage, without dividing by the total of 80 first. Giving 32.5% computes the percentage of chocolate cakes instead of fruit cakes: 26 ÷ 80 × 100 = 32.5%. Giving 42.5% computes the percentage of sponge cakes instead of fruit cakes: 34 ÷ 80 × 100 = 42.5%.
(c) Yes — with an estimate of 320, above the 250 limit. — Method: scale the sample proportion up to the whole batch to get an estimate, then compare that estimate with the 250 limit to reach a decision. Working: in the sample, 4 out of 50 boards are faulty, a proportion of 4 ÷ 50 = 0.08. Applying that proportion to the batch of 4,000 gives an estimate of 0.08 × 4000 = 320 faulty boards. Since 320 is more than 250, the factory should scrap the batch. Inverting the proportion, 50 ÷ 4 = 12.5, and treating that as a percentage of the batch, 12.5% × 4000 = 500, still gives 'yes' but from the wrong fraction, so it overstates the estimate. Comparing the raw number of faulty boards found in the sample, 4, directly with the 250 limit skips the scaling up to the batch altogether, and 4 is nowhere near 250, so that route wrongly says 'no'. Dividing the batch by the sample size, 4000 ÷ 50 = 80, finds how many samples of 50 fit into the batch but stops before multiplying by the 4 faulty boards found, so it also wrongly says 'no'. Always find the proportion in the sample first, scale it up to the whole batch, and only then compare the estimate with the limit given.

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