Less common Papers 1 & 2

Statistics: Averages, Spread, Histograms and Correlation

Summarising a data set with one average plus one measure of spread, then reading them back off charts. The TMUA rewards speed here: histogram frequency is AREA not height, the median comes from a position, and every comparison pairs a centre with a spread.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • Mean = (sum of all values) / (number of values); for a frequency table mean = (sum of f*x) / (sum of f). It uses every value, so a single outlier drags it noticeably.
  • Median is the middle value of ORDERED data, sitting at position (n+1)/2; if n is even you average the two middle values. It is resistant to outliers.
  • Mode = most frequent value (the modal class for grouped data); it is the only average usable for non-numerical or categorical data.
  • Spread: range = largest - smallest uses only two values, while the interquartile range IQR = Q3 - Q1 measures the middle 50% and ignores extremes.
  • Grouped data gives only an ESTIMATED mean: use class midpoints as x in (sum of f*x) / (sum of f); the original values are lost so it cannot be exact.
  • In a histogram the AREA of each bar equals frequency, so bar height = frequency density = frequency / class width. Use this whenever class widths are unequal.
  • Scatter graphs show correlation: positive (both rise together), negative (one rises as the other falls), or none. Correlation is not causation, and extrapolating a line of best fit past the data is unreliable.
  • To compare two data sets, quote one average AND one spread together: a higher mean with a larger range can still be the less consistent set.

Diagram

xy24682468line of best fit
Positive correlation: as one variable increases so does the other; a line of best fit summarises the trend.

Formulae

mean = (sum of f*x) / (sum of f)

Mean from a frequency table; for grouped data use class midpoints for x, giving an estimate.

median position = (n+1)/2

Locating the median in an ordered list of n values; an even n gives a half-integer, so average the two middle values.

frequency density = frequency / class width

Finding the height of a histogram bar, especially when class widths are unequal.

frequency = frequency density × class width

Recovering a class frequency from a histogram (the area of the bar).

IQR = Q3 - Q1

Measuring spread while ignoring outliers; comparing the consistency of two data sets.

range = largest - smallest

A quick, crude measure of total spread when no robust measure is needed.

Definitions

Frequency density
Frequency divided by class width; it is the height of a histogram bar, chosen so that the bar's AREA equals its frequency.
Interquartile range (IQR)
Q3 - Q1, the range of the central 50% of ordered data; unaffected by extreme values, so it is a robust measure of spread.
Modal class
In grouped data, the class with the highest frequency (for equal widths) or the highest frequency density (for unequal widths).
Correlation
The strength and direction of a linear relationship between two variables on a scatter graph: positive, negative, or none. It does not by itself prove causation.

Worked examples

1

Twenty customers' browsing times (minutes) are grouped: 0≤t<10 has 4 people, 10≤t<20 has 5, 20≤t<30 has 8, 30≤t<40 has 3. Estimate the mean time, and state which class contains the median.

  1. 1

    Use class midpoints:

    x = 5, 15, 25, 35
  2. 2

    Weight by frequency:

    f*x = 20, 75, 200, 105
  3. 3

    Add these up:

    sum f*x = 400
  4. 4

    Total frequency:

    n = 20
  5. 5

    Divide to estimate:

    mean = 400/20 = 20
  6. 6

    Median position:

    n/2 = 10th value
  7. 7

    Cumulate the frequencies:

    cf = 4, 9, 17, 20
  8. 8

    The running total passes 10 within the third class, so the median lies in 20≤t<30.

Answer: Estimated mean = 20 minutes; the median lies in the class 20≤t<30.

2

In a histogram, class A covers 20≤x<40 with frequency density 3, and class B covers 40≤x<50 with frequency density 5. Find the frequency in each class and say which class contains more data values.

  1. 1

    Class A width:

    wA = 40 - 20 = 20
  2. 2
    Frequency = density × width:
    fA = 3 × 20 = 60
  3. 3

    Class B width:

    wB = 50 - 40 = 10
  4. 4

    Same rule for B:

    fB = 5 × 10 = 50
  5. 5

    Since 60 > 50, class A holds more data despite having the shorter bar.

Answer: Class A has 60 values and class B has 50; class A contains more data, because frequency is the bar's AREA, not its height.

Common mistakes

  • ×Reading a histogram bar's HEIGHT as its frequency. Height is frequency density; frequency is the AREA = density × class width, which matters most when widths differ.
  • ×Taking the middle of an UNORDERED list, or muddling the position rule. Always sort first, then count to position (n+1)/2.
  • ×Treating the grouped-data mean as exact. Using midpoints only estimates the mean, because the individual values are no longer known.
  • ×Assuming correlation implies causation, or extending a line of best fit far beyond the plotted points (extrapolation) and trusting the result.
  • ×Comparing two data sets on an average alone (or a spread alone). A valid comparison always states both a centre and a spread.

No-calculator tips

  • ✓For a frequency-table mean, build the running total of f*x class by class and cancel common factors before the final division.
  • ✓Median position is (n+1)/2: an even n gives a half-integer (e.g. n=20 → 10.5), a direct signal to average the 10th and 11th values.
  • ✓If all class widths are equal you may rank histogram bars by height directly; only switch to frequency density when the widths differ.
  • ✓Adding a constant k to every value shifts the mean, median and mode by k but leaves range and IQR unchanged; multiplying by k scales all of them by k, so recompute nothing.
  • ✓Spot outliers first: one extreme value barely moves the median or mode but jumps the mean and range, which often settles a multiple-choice answer instantly.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.A data set has mean 12 and variance 9. Every value x is transformed using y = 3x - 5. What are the mean and variance of the new data set y?

  • A. mean 31, variance 27
  • B. mean 36, variance 81
  • C. mean 31, variance 81
  • D. mean 31, variance 76
  • E. mean 41, variance 81
Show answer

Answer: C — mean 31, variance 81

For y = ax + b the mean becomes a*mean + b = 3*12 - 5 = 31, and the variance becomes a2*variance = 32*9 = 81. The additive constant -5 does not affect spread, and the multiplier must be squared (not just 3*9 = 27).

Q2.On a histogram, the class 10 ≤ x < 30 has frequency density 4, and the class 30 ≤ x < 40 has frequency density 6. How many data values lie in these two classes in total?

  • A. 100
  • B. 200
  • C. 10
  • D. 140
Show answer

Answer: D — 140

Frequency = frequency density × class width. The widths are 20 and 10, giving 20*4 = 80 and 10*6 = 60, so the total is 140. Adding the densities (4 + 6 = 10) or assuming equal widths ignores that the class widths differ.

Q3.On a scatter graph the five points (1, 9), (3, 7), (5, 5), (7, 3), (9, 1) are plotted. What is the product moment correlation coefficient r for this data?

  • A. r = 1
  • B. r = -1
  • C. r = 0
  • D. r = -0.8
  • E. r = -1/2
Show answer

Answer: B — r = -1

Every point satisfies y = 10 - x, so they lie exactly on a straight line with negative gradient: the correlation is perfect and negative, giving r = -1. It is not +1 (that ignores the negative gradient) and not close-but-not-perfect since the fit is exact.

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