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Measures of location and spread questions
Two numbers summarise a thousand. One says where the data sits, the other says how far it strays. This lesson builds the mean, median and quartiles, then the standard deviation, and shows how coding moves them all in predictable ways.
6 original questions · 21 marks · the measures of location and spread notes · Statistics
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
For the data 4, 7, 7, 9, 13, write down the mean, median and mode.
Worked answer
Mean 40/5 = 8. Median 7, the middle of the ordered list. Mode 7, the most frequent value. B1 B1 B1 for the three measures. Three different answers from one small set, because the three measures genuinely measure different things. The 13 drags the mean above the median, which is what a high straggler always does.For 25 values, Σx = 450 and Σx2 = 8500. Find the mean and the standard deviation.
Worked answer
The mean is 450/25 = 18. The variance is 8500/25 − 182 = 340 − 324 = 16, so σ = √16 = 4. B1 for the mean, M1 for the variance formula, A1 for σ = 4. Take the mean of the squares and subtract the square of the mean, in that order. Reversing the two gives a negative variance, which is the instant sign that the formula has been written the wrong way round. Squaring is what stops the deviations above and below the mean cancelling to zero, and it is also why one wild value moves σ so much.Forty journey times are grouped: 8 in 0 ≤ t < 10, 14 in 10 ≤ t < 20, 12 in 20 ≤ t < 30, 6 in 30 ≤ t < 40. Use interpolation to estimate the median.
Worked answer
The median position is n/2 = 20. The first class holds 8, so the 20th value sits 12 into the second class, which holds 14 spread across a width of 10. Interpolating, the median is 10 + (12/14) × 10 = 18.6 minutes. M1 for the position n/2 = 20, M1 for a correct interpolation fraction, A1 for 18.6. The answer has to land inside 10 to 20, and if it does not, the running total has gone astray. For grouped continuous data Edexcel uses n/2 rather than (n + 1)/2, so use 20 here, not 20.5.Data are coded by y = (x − 100)/5, and the coded values have mean 4 and standard deviation 2. Find the mean and standard deviation of the original data.
Worked answer
Uncode in reverse, so x = 100 + 5y. The mean becomes 100 + 5 × 4 = 120 and the standard deviation becomes 5 × 2 = 10. B1 for the mean 120, M1 for scaling the standard deviation, A1 for 10. Subtracting 100 slid the whole data set along without stretching it, so only the division by 5 touched the spread. This is the general rule in miniature. Adding a constant shifts the mean and leaves the standard deviation alone; multiplying by a constant scales both.Two machines fill bags with mean mass 500 g each; machine A's masses have σ = 2 g and machine B's σ = 9 g. Compare the machines in context.
Worked answer
The centres agree, since both average 500 g. The spreads do not. Machine A's bags stay within a few grams of the target while machine B's stray four to five times as far, so machine A is much the more consistent filler. B1 for comparing the means, B1 for comparing the spreads, B1 for the context and units. Both sentences need the context and the units to score. Equal means say nothing about reliability, and it is the standard deviation that carries that part of the story.A set of 20 values has mean 15 and standard deviation 3. A further value of 36 is added to the set. Find the new mean and the new standard deviation.
Worked answer
Work back to the sums, because those are what combine. Σx = 20 × 15 = 300. For the squares, σ2 = Σx2/n − x̄2 rearranges to Σx2 = n(σ2 + x̄2) = 20(9 + 225) = 4680. Adding the new value: Σx becomes 300 + 36 = 336 and Σx2 becomes 4680 + 1296 = 5976, with n now 21. The new mean is 336/21 = 16. The new variance is 5976/21 − 162 = 284.57 − 256 = 28.57, so the new standard deviation is 5.35. M1 for recovering Σx, M1 for recovering Σx2, M1 for updating both sums, A1 for the new mean 16, M1 for the new variance, A1 for 5.35. Averaging the old standard deviation with anything is worth nothing, because standard deviations do not add. Only Σx and Σx2 do. A single value 7 standard deviations out has lifted the mean by 1 and nearly doubled the spread, which shows how little it takes to distort both.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise measures of location and spread one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.