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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- (a) 1,400 and 1,240, so combine the samples for one estimate — Method: scale each sample up to the whole stock, then use the fact that a larger sample gives a more reliable estimate than a smaller one. Working: the first sample gives 35 ÷ 50 = 0.7 and 0.7 × 2,000 = 1,400 paperbacks; the second gives 31 ÷ 50 = 0.62 and 0.62 × 2,000 = 1,240 paperbacks. Two random samples of the same size are expected to differ a little, so neither estimate is wrong. Putting the two together gives 35 + 31 = 66 paperbacks in 100 books, and 66 ÷ 100 = 0.66 with 0.66 × 2,000 = 1,320, an estimate resting on twice as many books as either volunteer checked. Answer: 1,400 and 1,240, so combine the samples for one estimate. The distractors: keeping 1,400 because it is larger picks an estimate by its size, when both samples held 50 books and neither has a stronger claim; saying a volunteer must have miscounted assumes two random samples ought to agree exactly, which is precisely what random sampling does not promise; 1,750 and 1,550 come from 35 × 50 = 1,750 and 31 × 50 = 1,550, multiplying each count by the size of the sample instead of scaling by 2,000 ÷ 50.
- (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.
- (a) 31.5 — Method: to estimate the mean from a histogram, first turn each bar into a frequency (frequency density × class width), then use mean = Σ(frequency × midpoint) ÷ Σfrequency, with the midpoint standing in for every value in that class. Working: the four classes have widths 20, 10, 20 and 20, so their frequencies are 1 × 20 = 20, 3 × 10 = 30, 2 × 20 = 40 and 0.5 × 20 = 10, which do add to the 100 vehicles stated. Their midpoints are 10, 25, 40 and 60, so Σfx = 20 × 10 + 30 × 25 + 40 × 40 + 10 × 60 = 200 + 750 + 1600 + 600 = 3150, and the mean is 3150 ÷ 100 = 31.5. Answer: the estimated mean speed is 31.5 mph. Watch which numbers you treat as the frequencies and which as the values: using the frequency densities themselves as the frequencies, without multiplying by the class widths first, gives 1 × 10 + 3 × 25 + 2 × 40 + 0.5 × 60 = 195 spread over 1 + 3 + 2 + 0.5 = 6.5, and 195 ÷ 6.5 = 30, a mean built from the wrong 'frequencies' altogether; averaging the four midpoints on their own, (10 + 25 + 40 + 60) ÷ 4 = 33.75, ignores how many vehicles are actually in each class; and using each class's lower boundary in place of its midpoint, 20 × 0 + 30 × 20 + 40 × 30 + 10 × 50 = 2300 and 2300 ÷ 100 = 23, systematically underestimates every class by roughly half its width.
- (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) Drawing 60 names at random from a list of all 1200 pupils — Method: a sample is random when every member of the population has the same chance of being chosen and nobody, including the pupils themselves, can influence who ends up in it; test each method against that. Working: drawing names from a list of all 1200 pupils gives each pupil the same chance, 60 out of 1200, whatever their year group, class or opinion, so the method is random. Answer: drawing 60 names at random from a list of all 1200 pupils. The distractors: asking the pupils who volunteer is self-selection, and the pupils with the strongest views volunteer first, so they decide the sample; asking the pupils nearest the door is convenience sampling, which reaches only those who happen to be in one place at one time; asking two Year 10 classes samples a cluster, so every pupil in the other year groups has no chance of being chosen at all.
- (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) 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.
- (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.
- (c) No, 90 cm is far outside the heights on the graph — Method: a line of best fit describes the trend only across the stretch of data it was drawn through; predicting beyond that stretch is extrapolation, and nothing in the data supports it. Working: the heights used to draw this line run from 150 cm to 180 cm, all of them Year 10 pupils, while 90 cm is 60 cm below the shortest of them and belongs to a two-year-old child, whose build follows no trend the graph has measured. Substituting anyway gives 0.9 × 90 − 85 = −4, a mass of −4 kg, which cannot exist. Answer: no, because 90 cm is far outside the heights on the graph. The distractors: saying a line of best fit cannot be used to predict at all throws away its main purpose, since a prediction made between the plotted values is perfectly sound; saying the line passes through all 20 points misdescribes a line of best fit, which is drawn to follow the trend of the points and will normally pass through few of them; saying the equation works for any value put into it treats an equation fitted to Year 10 heights as a law of nature, and the mass of −4 kg shows what that assumption produces.
- (c) 69 — Method: multiply the mean by the number of values to find the total, then subtract the total of the known values. Working: the total of all five scores is 68 × 5 = 340. The total of the four known scores is 55 + 62 + 74 + 80 = 271. The fifth score is 340 − 271 = 69. Subtracting the other way round, 271 − 340 = −69, gives the right size answer with the wrong sign. Guessing that the missing score simply equals the mean, 68, ignores that the four known scores are not themselves centred on 68. Multiplying the mean by 4 instead of 5, 68 × 4 = 272, then 272 − 271 = 1, undercounts how many scores there are. Always multiply the mean by the TOTAL number of values before subtracting.
- (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.
- (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.
- (d) 4, 3, 2, 6 — Method: frequency density = frequency ÷ class width for each class in turn; do not assume the classes are all the same width. Working: the four classes have widths 10 − 0 = 10, 30 − 10 = 20, 45 − 30 = 15 and 50 − 45 = 5. Dividing each frequency by its own width gives 40 ÷ 10 = 4, 60 ÷ 20 = 3, 30 ÷ 15 = 2 and 30 ÷ 5 = 6. Answer: the frequency densities, in order, are 4, 3, 2 and 6. Watch the width of each class separately: treating the last class as if it were also 10 units wide, like the first, gives 30 ÷ 10 = 3 instead of 30 ÷ 5 = 6 — the classes here are deliberately unequal, so no width can be borrowed from another class; dividing the width by the frequency instead of the frequency by the width for the third class gives 15 ÷ 30 = 0.5 in place of 2, the formula the wrong way round; and reading the frequency column straight off the table, 40, 60, 30, 30, skips the division by width altogether and reports how many fish are in each class rather than how densely packed each bar is.
- (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.
- (a) Leeds has a higher median and a greater range than York. — In order, Leeds's temperatures are 14, 16, 18, 19 and 23, so the median is the middle value, 18, and the range is 23 − 14 = 9. York's temperatures in order are 15, 17, 17, 18 and 18, so the median is 17, and the range is 18 − 15 = 3. Since 18 is higher than 17, and 9 is greater than 3, Leeds has both the higher median and the greater range. Choosing 'Leeds has a higher median but a smaller range than York' gets the median comparison right but the range comparison backwards — Leeds's range of 9 is actually greater than York's range of 3. Choosing 'York has a higher median and a greater range than Leeds' reverses both comparisons. Choosing 'York has a higher median but a smaller range than Leeds' reverses the median comparison; York's median of 17 is lower than Leeds's 18, even though it is correct that York's range is the smaller one.
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