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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- (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.
- (b) £11.00 — 1.5 × 6 = 9, and 9 + 2 = 11, so the estimated cost is £11.00. Choosing £9.00 stops after 1.5 × 6 = 9 and forgets to add the £2. Choosing £12.00 adds the mass and the constant first and then multiplies: 6 + 2 = 8, and 8 × 1.5 = 12.00. Choosing £13.50 swaps the gradient and the intercept, using y = 2x + 1.5 instead: 2 × 6 = 12, and 12 + 1.5 = 13.50.
- (d) Only internet users reach the website; others are excluded. — Method: a sample is biased when it systematically leaves out part of the population, or systematically over-represents another part. Working: anyone without internet access, or who does not visit the council's website, has NO chance of being included — the sample is drawn only from internet-using residents, which is not the whole town. Saying too many people might respond because the survey is free confuses bias with sample size — bias is about who CAN be reached, not how many respond. Saying people might lie describes a different problem, response honesty, not who was sampled in the first place. Saying online surveys cannot be anonymous is not a reason connected to bias at all. A sample is biased when part of the population has no chance of being included, whatever the reason for that.
- (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.
- (b) 55 — Method: count the classes that lie wholly above 8 minutes, then use linear interpolation for the class that 8 cuts through, assuming the delays in that class are spread evenly. Working: the class 10 ≤ d < 20 lies wholly above 8 and holds 25 buses; the value 8 lies in the class 5 ≤ d < 10, which is 5 minutes wide and holds 75 buses, and the part above 8 runs from 8 to 10, a width of 2, so the estimated share is (2 ÷ 5) × 75 = 30 buses; the estimate is 30 + 25 = 55. Answer: about 55 refunds. The distractors: 100 comes from adding the whole of the class 5 ≤ d < 10, 75 + 25, and so refunding buses only 5 minutes late; 25 comes from using only the class 10 ≤ d < 20 and ignoring the part class that 8 minutes cuts through; 70 comes from taking the part of the class from 5 up to 8 instead of from 8 up to 10, giving (3 ÷ 5) × 75 = 45 and then 45 + 25.
- (a) 30 — Method: an outlier is a value that lies far away from the pattern set by the rest of the data, so compare each value with the group the others form. Working: five of the counts, 5, 7, 8, 9 and 11, lie within 6 of one another and the steps between them are 2, 1, 1 and 2; the remaining count of 30 is 19 above the nearest of them, so it is the value that does not belong to the pattern. Answer: 30. The distractors: 5 comes from picking the smallest value, on the idea that the odd one out must be at the bottom of the list; 11 comes from ordering the data and stopping one value short, taking the largest of the counts that sit close together; 8.5 comes from working out the median, (8 + 9) ÷ 2, and giving a measure of centre where a value standing apart was asked for.
- (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%.
- (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.
- (c) 28 — Method: multiply the number of whole symbols by the value of one symbol, then add the value of any half symbol shown. Working: 3 whole symbols represent 3 × 8 = 24 cars. The half symbol represents 4 cars. Total cars sold in March = 24 + 4 = 28. Leaving out the half symbol, 3 × 8 = 24, undercounts by exactly the value of that half symbol. Treating the half symbol as if it were a full symbol, 4 × 8 = 32, overcounts because it doubles the value the half symbol is worth. Giving 3.5 reports the number of symbols shown, not the number of cars they represent — the key still needs to be applied. Always apply the key to every symbol shown, including a half symbol, rather than reading off the symbol count itself.
- (d) 16 — Method: rearrange interquartile range = upper quartile − lower quartile to make the lower quartile the subject: lower quartile = upper quartile − interquartile range, then check the answer sits in the right place in the list. Working: there are 11 values, so 11 + 1 = 12; the upper quartile sits at position 3 × 12 ÷ 4 = 9, which is 35, and x sits at position 12 ÷ 4 = 3, which is the lower quartile. So x = 35 − 19 = 16, and 16 does sit between the 2nd value, 9, and the 4th value, 17, as it should. Answer: x = 16. Watch how you rearrange and where you count to: adding instead of subtracting, 35 + 19 = 54, treats the interquartile range as something added on rather than a gap taken away; subtracting in the wrong order, 19 − 35 = −16, finds the right two numbers but flips the sign; and counting to the 8th value instead of the 9th treats 31 as the upper quartile, giving 31 − 19 = 12, one position short of where the upper quartile actually sits.
- (c) 33 — Method: multiply the mean by the number of tests to get the total marks, then subtract the marks that are already known. Working: four tests with a mean of 29 give a total of 29 × 4 = 116 marks; the first three marks total 31 + 26 + 26 = 83; so the fourth mark is 116 − 83 = 33. Answer: 33, and checking, (31 + 26 + 26 + 33) ÷ 4 = 116 ÷ 4 = 29. The distractors: 116 comes from stopping at the total for all four tests; 29 comes from assuming the missing mark must be the mean itself; 4 comes from multiplying the mean by 3, the number of marks given, leaving 87 − 83 = 4.
- (d) 23.57 °C — Method: add all seven temperatures, divide by the number of readings and round only at the end. Working: 22 + 24 + 23 + 25 + 26 + 21 + 24 = 165, and 165 ÷ 7 = 23.5714…, which rounds to 23.57 to 2 decimal places. Answer: 23.57 °C. The distractors: 24 °C comes from writing down the mode, the only temperature recorded twice, instead of the mean; 27.5 °C comes from dividing the total by 6 instead of by the 7 days recorded; 5 °C comes from working out the range, 26 − 21, which is a measure of spread and not an average.
- (a) x = 0 gives y = −20: a negative number sold — The y-intercept is the value the line predicts when x = 0: y = 3 × 0 − 20 = −20. A kiosk cannot sell a negative number of ice creams, so this is not a sensible estimate. The 3 in the equation is the gradient, not the intercept, so an option claiming x = 0 gives y = 3 has swapped the two numbers around — substituting x = 0 makes the 3x term equal 0, leaving −20, not 3. The danger of extrapolating to very high temperatures is a real issue with this line, but it is a different issue from the y-intercept, so it does not answer this question. And whether x = 0 could occur on a trading day is beside the point: the model still makes that prediction, and it is the prediction itself, −20, that is impossible.
- (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.
- (c) Class X has the higher median and the wider spread — Method: compare the two box plots statistic by statistic — median for location, and the interquartile range for spread — checking the true value of each rather than assuming a pattern. Working: Class X's median is 60 and Class Y's is 58, so Class X's median is the higher one. Class X's interquartile range is 70 − 45 = 25 and Class Y's is 65 − 50 = 15 (and the ranges follow the same order: 95 − 20 = 75 against 80 − 35 = 45), so Class X also has the wider spread. Answer: Class X has both the higher median and the wider spread. Watch that each half of a compound statement is checked separately: claiming Class Y has the higher median and the wider spread gets both comparisons backwards; claiming Class X has the higher median but the narrower spread keeps the median right while reading the spread the wrong way round; and claiming Class Y has the higher median but the narrower spread swaps the median comparison while getting the spread right.
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