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Statistics

Mean, median, mode and range (including grouped estimates), histograms and frequency density, cumulative frequency, quartiles, scatter graphs.

Spec M6.1-M6.4Section 1 Part 2: Higher GCSE mathematics11 questions3 real past questions
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1Statistics · spec M6.3 · Paper 1 style · ★☆☆

The mean of five numbers is \(12\). When a sixth number is added the mean becomes \(14\). What is the sixth number?

  1. A\(24\)
  2. B\(84\)
  3. C\(20\)
  4. D\(26\)
  5. E\(16\)
  6. F\(14\)
Worked solution

Total goes from \(60\) to \(84\): the sixth number is \(24\).

Full method — how and why
Means questions are almost always really about totals. Convert each mean into a total before doing anything else.
  1. Original total \(= 5 \times 12 = 60\).
    mean \(=\dfrac{\text{total}}{\text{how many}}\), so total \(=\) mean \(\times\) how many.
  2. New total \(= 6 \times 14 = 84\).
    There are now six numbers, and their mean is \(14\).
  3. The sixth number is the difference: \(84 - 60 = 24\).
    Nothing else changed, so the whole increase in the total is the new number.
Where it goes wrong
  • Answering \(14\) - the new mean, not the new number.
  • Answering \(84\), the new total rather than the added value.
  • Thinking the new number must be \(2\) more than the old mean; the mean rose by \(2\) for all six, so the newcomer must carry \(6 \times 2 = 12\) extra above the old mean of \(12\), giving \(24\).
Takeaway Turn means into totals; the added value is the change in total.
Answer A · Statistics (M6.3) · question STA-01
2Statistics · spec M6.3 · Paper 1 style · ★☆☆

A histogram has a bar for the class \(10 \leq x < 15\) of height (frequency density) \(3.2\) and a bar for \(15 \leq x < 25\) of height \(1.5\). What is the total frequency of these two classes?

  1. A\(15\)
  2. B\(23.5\)
  3. C\(31\)
  4. D\(4.7\)
  5. E\(16\)
  6. F\(47\)
Worked solution

Frequency \(=\) density \(\times\) width: \(3.2 \times 5 + 1.5 \times 10 = 16 + 15 = 31\).

Full method — how and why
On a histogram the area of a bar is the frequency, not its height. Height is frequency density.
  1. Read the first bar: width \(15 - 10 = 5\), height \(3.2\).
    Class widths on a histogram are usually unequal - that is why density is used.
  2. Frequency \(=\) density \(\times\) width \(= 3.2 \times 5 = 16\).
    Rearranging density \(=\dfrac{\text{frequency}}{\text{width}}\).
  3. Second bar: width \(25 - 15 = 10\), so frequency \(= 1.5 \times 10 = 15\).
    Same rule, wider class.
  4. Add: \(16 + 15 = 31\).
    Total frequency of the two classes.
Where it goes wrong
  • Adding the heights: \(3.2 + 1.5 = 4.7\) - an offered option.
  • Using the same width for both bars; the second is twice as wide even though it is shorter.
Takeaway Histogram: frequency \(=\) density \(\times\) class width (area of the bar).
Answer C · Statistics (M6.3) · question STA-02
3Statistics · spec M6.3 · Paper 1 style · ★☆☆

The numbers \(3, 7, x, 12, 18\) (with \(7 \leq x \leq 12\)) have mean equal to median. Find \(x\).

  1. A\(12\)
  2. B\(7\)
  3. C\(10\)
  4. D\(8\)
  5. E\(11\)
  6. F\(9\)
Worked solution

The median is \(x\) and the mean is \(\frac{40 + x}{5}\). Setting them equal: \(40 + x = 5x\), so \(x = 10\).

Full method — how and why
Write expressions for the mean and the median, then equate. The given range for \(x\) tells you where it sits in the order.
  1. Since \(7 \leq x \leq 12\), the sorted list is \(3, 7, x, 12, 18\), so the median is \(x\).
    With five values the median is the third; the constraint is there precisely to fix \(x\)'s position.
  2. Mean \(= \dfrac{3 + 7 + x + 12 + 18}{5} = \dfrac{40 + x}{5}\).
    Sum the known values first: \(3+7+12+18 = 40\).
  3. Set mean \(=\) median: \(\dfrac{40+x}{5} = x\).
    That is the condition stated.
  4. Solve: \(40 + x = 5x \Rightarrow 4x = 40 \Rightarrow x = 10\).
    And \(10\) does lie in \([7, 12]\), so the assumed ordering holds.
Where it goes wrong
  • Taking the median to be \(12\) (the middle of the numbers as originally listed without \(x\)).
  • Forgetting to check that the answer is consistent with the assumed position in the order.
Takeaway Order first, then write the median; always check the solution respects the ordering you assumed.
Answer C · Statistics (M6.3) · question STA-03
4Statistics · spec M6.3 · Paper 1 style · ★☆☆

The table shows grouped data: \(0 \leq t < 10\): frequency \(4\); \(10 \leq t < 20\): frequency \(10\); \(20 \leq t < 40\): frequency \(6\). Estimate the mean.

  1. A\(16\)
  2. B\(18\)
  3. C\(20\)
  4. D\(15\)
  5. E\(22\)
  6. F\(17.5\)
Worked solution

Use the midpoints \(5, 15, 30\): \(\dfrac{4\cdot5 + 10\cdot15 + 6\cdot30}{20} = \dfrac{20 + 150 + 180}{20} = \dfrac{350}{20} = 17.5\).

Full method — how and why
Grouped data: you cannot know the individual values, so estimate using class midpoints.
  1. Find midpoints: \(\frac{0+10}{2} = 5\), \(\frac{10+20}{2} = 15\), \(\frac{20+40}{2} = 30\).
    The midpoint is the best single representative of a class.
  2. Multiply each by its frequency: \(4(5) = 20\), \(10(15) = 150\), \(6(30) = 180\).
    This estimates the total of all the values in each class.
  3. Total \(= 350\); total frequency \(= 4 + 10 + 6 = 20\).
    Add both columns.
  4. Mean \(= \dfrac{350}{20} = 17.5\).
    Divide estimated total by the number of observations.
Where it goes wrong
  • Using the midpoint \(25\) for the last class by assuming equal widths - it runs from \(20\) to \(40\), so its midpoint is \(30\).
  • Dividing by \(3\) (the number of classes) instead of \(20\).
Takeaway Grouped mean \(\approx \dfrac{\sum f x_{\text{mid}}}{\sum f}\) - and read each class width individually.
Answer F · Statistics (M6.3) · question STA-04
5Statistics · spec M6.4 · Paper 2 style · ★☆☆

Which of the following statements about a scatter graph showing strong positive correlation between hours of revision and exam mark is/are valid?

  1. IStudents who revised more tended to score higher.
  2. IIRevising more causes a higher mark.
  3. IIIA line of best fit could be used to estimate the mark of a student who revised for a number of hours within the range of the data.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

Correlation describes a tendency (I) and supports interpolation within the data range (III), but does not by itself establish causation (II).

Full method — how and why
A correlation-and-causation statement test. Decide what correlation does and does not licence.
  1. I: strong positive correlation means higher revision went with higher marks. True - that is exactly what correlation describes.
    'Tended to' is a statement about the trend, not about every student.
  2. II: correlation does not establish causation. False.
    A hidden variable (motivation, prior ability) could drive both, so the data alone cannot prove cause.
  3. III: using a line of best fit within the data range is interpolation. True.
    Interpolation is legitimate; extrapolation beyond the data is what is unreliable.
  4. So I and III only.
    Match to the list.
Where it goes wrong
  • Accepting II because the causal story sounds plausible - plausibility is not evidence.
  • Rejecting III by confusing interpolation with extrapolation.
Takeaway Correlation licenses description and interpolation; it never licenses a causal claim on its own.
Answer F · Statistics (M6.4) · question STA-05
6Statistics · spec M6.3 · Paper 1 style · ★☆☆

A set of \(n\) numbers has mean \(m\). Each number is doubled and then \(3\) is added. Which of the following is the new mean, and which the new range (if the original range was \(R\))?

  1. Amean \(2m + 3\), range \(2R\)
  2. Bmean \(2m\), range \(2R\)
  3. Cmean \(m + 3\), range \(R\)
  4. Dmean \(2m + 3\), range \(2R + 3\)
  5. Emean \(2m + 3\), range \(R\)
Worked solution

Mean transforms with the data: \(2m + 3\). The range is unaffected by adding a constant but doubles with the scaling: \(2R\).

Full method — how and why
Coding: see separately how a linear transformation affects a measure of location (the mean) and a measure of spread (the range).
  1. Each value \(x\) becomes \(2x + 3\).
    Read the transformation carefully: double, then add.
  2. The mean follows the same transformation: new mean \(= 2m + 3\).
    Because \(\overline{2x+3} = 2\bar x + 3\) - the mean is linear.
  3. The range is a difference: \((2x_{\max}+3) - (2x_{\min}+3) = 2(x_{\max} - x_{\min}) = 2R\).
    The \(+3\) cancels in the subtraction; only the scaling survives.
  4. So mean \(2m+3\), range \(2R\).
    Location shifts and scales; spread only scales.
Where it goes wrong
  • Giving range \(2R + 3\) - the offered trap; adding a constant to every value shifts the data but does not stretch it.
  • Leaving the range unchanged as \(R\), forgetting the doubling.
Takeaway Under \(x \to ax + b\): mean \(\to a\bar x + b\), but spread (range, IQR, SD) \(\to |a| \times\) spread - the \(+b\) never affects spread.
Answer A · Statistics (M6.3) · question STA-06
7Statistics · spec M6.3 · Paper 1 style · ★☆☆

A cumulative frequency graph for \(80\) observations passes through \((20, 20)\), \((30, 40)\) and \((45, 60)\). Estimate the interquartile range.

  1. A\(40\)
  2. B\(10\)
  3. C\(20\)
  4. D\(25\)
  5. E\(15\)
  6. F\(30\)
Worked solution

Lower quartile at cumulative frequency \(20\): value \(20\). Upper quartile at \(60\): value \(45\). IQR \(= 25\).

Full method — how and why
On a cumulative frequency graph, quartiles are read by going in from the cumulative frequency axis at \(\frac n4\), \(\frac n2\), \(\frac{3n}{4}\).
  1. With \(n = 80\): the lower quartile is at cumulative frequency \(\frac{80}{4} = 20\), the upper at \(\frac{3 \times 80}{4} = 60\).
    Cumulative frequency graphs use \(\frac n4\) and \(\frac{3n}{4}\) directly (no \(+1\), unlike a raw list).
  2. The point \((20, 20)\) gives \(Q_1 = 20\).
    Read across at CF \(= 20\) and down to the value axis.
  3. The point \((45, 60)\) gives \(Q_3 = 45\).
    Read across at CF \(= 60\).
  4. IQR \(= Q_3 - Q_1 = 45 - 20 = 25\).
    The interquartile range is a difference of values, not of frequencies.
Where it goes wrong
  • Subtracting the cumulative frequencies, \(60 - 20 = 40\) - an offered option.
  • Using the median point \((30, 40)\) as a quartile.
Takeaway Go in from the frequency axis at \(\frac n4\) and \(\frac{3n}{4}\), read off values, then subtract the values.
Answer D · Statistics (M6.3) · question STA-08
8Statistics · spec M6.3 · Paper 1 style · ★☆☆

Class A has \(20\) students with mean mark \(60\); class B has \(30\) students with mean mark \(70\). What is the mean mark of all \(50\) students?

  1. A\(68\)
  2. B\(65\)
  3. C\(62\)
  4. D\(67\)
  5. E\(64\)
  6. F\(66\)
Worked solution

\(\frac{1200 + 2100}{50} = \frac{3300}{50} = 66\).

Full method — how and why
A combined mean is a weighted average: go back to totals, since the classes are different sizes.
  1. Class A total \(= 20 \times 60 = 1200\).
    Total \(=\) mean \(\times\) number.
  2. Class B total \(= 30 \times 70 = 2100\).
    Same again for the larger class.
  3. Combined mean \(= \dfrac{1200 + 2100}{50} = \dfrac{3300}{50} = 66\).
    Divide the grand total by the grand number of students.
Where it goes wrong
  • Averaging the means: \(\frac{60+70}{2} = 65\) - an offered option, and wrong because class B has more students.
  • A quick check: the answer must lie closer to \(70\) than to \(60\), which \(66\) does.
Takeaway Never average averages; weight by group size - or just work with totals.
Answer F · Statistics (M6.3) · question STA-09
9Statistics · spec M6.3 · Paper 1 style · ★☆☆

For which of the following data sets is the mean greater than the median?

  1. A\(1, 2, 3, 4, 5\)
  2. B\(5, 5, 5, 5, 5\)
  3. C\(2, 4, 6, 8, 10\)
  4. D\(1, 2, 3, 4, 20\)
  5. E\(1, 10, 11, 12, 13\)
Worked solution

A large outlier pulls the mean above the median: mean \(6 >\) median \(3\). In B, D, E the mean equals the median; in C the mean \(9.4 <\) median \(11\).

Full method — how and why
The mean is pulled by extreme values; the median is not. Look for skew rather than computing everything.
  1. Check \(1, 2, 3, 4, 20\): median \(= 3\), mean \(= \frac{30}{5} = 6\). Mean \(>\) median. Yes.
    The single large outlier \(20\) drags the mean far above the middle.
  2. \(1,2,3,4,5\) and \(2,4,6,8,10\) and \(5,5,5,5,5\) are symmetric, so mean \(=\) median.
    For symmetric data the two coincide - no need to compute.
  3. \(1, 10, 11, 12, 13\): median \(= 11\), mean \(= \frac{47}{5} = 9.4\), so mean \(<\) median.
    Here the outlier is small, so it drags the mean down - the mirror image.
  4. So only \(1, 2, 3, 4, 20\) works.
    One high outlier gives positive skew.
Where it goes wrong
  • Assuming any outlier raises the mean - a low outlier lowers it, as in \(1, 10, 11, 12, 13\).
  • Computing all five means when symmetry settles three of them instantly.
Takeaway Mean \(>\) median signals a tail to the right (a high outlier); mean \(<\) median signals a tail to the left.
Answer D · Statistics (M6.3) · question STA-10
10Statistics · spec M6.3 · Paper 1 style · ★★☆

The median of \(7\) distinct integers is \(10\) and the mean is \(12\). The smallest number is increased by \(20\). Which of the following must be true?

  1. AThe mean and median both increase.
  2. BThe median increases and the mean does not change.
  3. CThe mean increases and the median does not change.
  4. DThe mean increases and the median may or may not change.
  5. ENeither the mean nor the median changes.
Worked solution

The mean rises by \(\frac{20}{7}\). The smallest number plus \(20\) could exceed the median (e.g. if the smallest is \(1\) and it becomes \(21\) the middle value changes), or it could stay below it. So the median may or may not change.

Full method — how and why
A 'must be true' question. Test each claim by trying to construct a counterexample.
  1. The mean: the total rises by \(20\), so the mean rises by \(\frac{20}{7}\) - it definitely increases.
    Nothing else changed, so the total change is \(+20\) across \(7\) values.
  2. The median: try the smallest \(= 1\) becoming \(21\). That value jumps past the middle, so the values shift along and the median changes.
    Moving a value from below the median to above it pushes the middle value up one place.
  3. Now try the smallest \(= -15\) becoming \(5\), still below the median \(10\). The ordering of the middle is untouched, so the median is unchanged.
    A value that stays on the same side of the median cannot move it.
  4. Both are possible, so the median may or may not change while the mean definitely rises.
    That is precisely the statement offered.
Where it goes wrong
  • Claiming the median never changes - true only if the increased value stays below the median.
  • Claiming both must increase, which would need the smallest to always overtake the middle.
Takeaway 'Must be true' means no counterexample exists; hunt for one before agreeing.
Answer D · Statistics (M6.3) · question STA-07
11Statistics · spec M6.3 · Paper 2 style · ★★☆

A set of five distinct positive integers has median \(10\) and mean \(10\). Which of the following statements must be true?

  1. IThe largest value is at least \(12\).
  2. IIThe smallest value is at most \(8\).
  3. IIIThe range is at least \(4\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

The two values above the median are distinct integers greater than \(10\), so at least \(11\) and \(12\) (I); similarly the two below are at most \(9\) and \(8\) (II); hence the range is at least \(12 - 8 = 4\) (III).

Full method — how and why
Extremal reasoning with distinct integers: the distinctness forces gaps of at least \(1\) on each side of the median.
  1. Order the five values. The median (third value) is \(10\).
    Five values means the middle one is the third.
  2. The two above are distinct integers greater than \(10\), so at least \(11\) and \(12\). Hence the largest is \(\geq 12\). I true.
    Distinct integers above \(10\) must be \(11\) or more, and two of them cannot both be \(11\).
  3. Symmetrically, the two below are at most \(9\) and \(8\), so the smallest is \(\leq 8\). II true.
    Same argument downwards.
  4. Range \(\geq 12-8 = 4\). III true.
    Combining the two bounds. (The mean being \(10\) is consistent - e.g. \(8,9,10,11,12\) - but is not needed for any of the three claims.)
  5. So all three.
    Match to the list.
Where it goes wrong
  • Forgetting 'distinct' and allowing two values equal to \(11\), which would break I.
  • Assuming the mean condition must be used; here it is extra information.
Takeaway Distinct integers force minimum gaps: \(k\) values above a median \(m\) reach at least \(m+k\).
Answer H · Statistics (M6.3) · question X2-28

How this topic has been examined

Real TMUA questions on Statistics, newest first. Links open the official paper at the question and the online marking page.

2020 · Paper 2Question 17means and medians of split setsquestion papersit online
October 2018 · Paper 1Question 17exchanging data between two setsquestion papersit online
Specimen papers · Paper 2Question 18five numbers with mean 0 and range 20question papersit online