A-Level Measures of location and spread Practice Questions

Free A-Level Measures of location and spread practice questions with full step-by-step worked solutions. Covers mean, raw data, median, mode. Practise exam-style problems and check your method.

meanraw datamedianmoderangefrequency table
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The daily rainfall (in mm) over five days was 4,8,6,10,24, 8, 6, 10, 2. Find the mean daily rainfall.
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Worked solution

  1. Add up all the values

    ∑x=4+8+6+10+2=30\sum x = 4+8+6+10+2 = 30

    Find the total of the data values.

  2. Count how many values there are

    n=5n = 5

    There are 55 values in the list.

  3. Divide the total by the number of values

    xˉ=∑xn=305=6\bar{x} = \frac{\sum x}{n} = \frac{30}{5} = 6

    The mean is the total divided by how many values there are.

Answer
xˉ=6 mm\bar{x}=6\text{ mm}
Question 2
2 markseasy
How is the range\textbf{range} of a data set calculated?
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Worked solution

  1. Recall what range measures

    range=a measure of spread\text{range} = \text{a measure of spread}

    The range describes how spread out the data are.

  2. Use the extreme values

    range=max⁡−min⁡\text{range} = \max - \min

    Only the largest and smallest values are needed.

  3. Select the correct method

    max⁡−min⁡\max - \min

    Subtract the smallest value from the largest.

Answer
Subtract the smallest value from the largest value\text{Subtract the smallest value from the largest value}
Question 3
3 marksintermediate
Two classes sit the same test. Class A has mean 6060 and standard deviation 44; Class B has mean 6060 and standard deviation 1212. Which statement is correct?
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Worked solution

  1. Compare the means

    xˉA=xˉB=60\bar{x}_A = \bar{x}_B = 60

    Both classes have the same average mark.

  2. Compare the standard deviations

    σB=12>σA=4\sigma_B = 12 > \sigma_A = 4

    A larger standard deviation means more variation.

  3. Interpret the spread

    Class B more spread out\text{Class B more spread out}

    Class B's marks are more widely dispersed about the mean.

  4. Test the first option against the reasoning

    option A: ✓\text{option A}:\ \checkmark

    Option A agrees with every step of the argument.

  5. Test option B

    option B: ×\text{option B}:\ \times

    Option B does not follow from the definition and is rejected.

  6. State the final choice

    ⇒ option A\Rightarrow\ \boxed{\text{option A}}

    Select the statement consistent with all the reasoning.

Answer
Class B’s marks are more spread out than Class A’s\text{Class B's marks are more spread out than Class A's}
Question 4
5 markshard
An outlier that is much larger than the rest of the data is added to a sample. This will generally:
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Worked solution

  1. Effect on the mean

    xˉ ↑\bar{x}\ \uparrow

    A very large value increases the total, so the mean rises.

  2. Effect on the spread

    deviations from the mean ↑\text{deviations from the mean }\uparrow

    The extreme value is far from the mean, increasing the spread.

  3. Combine the effects

    xˉ↑, σ↑\bar{x}\uparrow,\ \sigma\uparrow

    Both the mean and standard deviation increase.

  4. Test the first option against the reasoning

    option A: ✓\text{option A}:\ \checkmark

    Option A agrees with every step of the argument.

  5. Test option B

    option B: ×\text{option B}:\ \times

    Option B does not follow from the definition and is rejected.

  6. Test option C

    option C: ×\text{option C}:\ \times

    Option C confuses a measure of location with a measure of spread and is rejected.

  7. Test option D

    option D: ×\text{option D}:\ \times

    Option D would only hold in a special case that does not apply here.

  8. Test option E

    option E: ×\text{option E}:\ \times

    Option E reverses the correct relationship and is rejected.

  9. Compare the surviving options

    only option A remains\text{only option A remains}

    All the distractors have been eliminated.

  10. State the final choice

    ⇒ option A\Rightarrow\ \boxed{\text{option A}}

    Select the statement consistent with all the reasoning.

Answer
Increase the mean and increase the standard deviation\text{Increase the mean and increase the standard deviation}
Question 5
8 markschallenging
A charity records donation amounts; most are small but a handful are extremely large. To report both a typical donation and a measure of spread that is least distorted by the large values, which pair of statistics is most appropriate?
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Worked solution

  1. Choose a robust measure of location

    median\text{median}

    The median is not pulled up by the few very large donations.

  2. Choose a robust measure of spread

    IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

    The interquartile range ignores the extreme tails.

  3. Combine the choices

    median and IQR\text{median and IQR}

    Both resist the influence of outliers.

  4. Test the first option against the reasoning

    option A: ✓\text{option A}:\ \checkmark

    Option A agrees with every step of the argument.

  5. Test option B

    option B: ×\text{option B}:\ \times

    Option B does not follow from the definition and is rejected.

  6. Test option C

    option C: ×\text{option C}:\ \times

    Option C confuses a measure of location with a measure of spread and is rejected.

  7. Test option D

    option D: ×\text{option D}:\ \times

    Option D would only hold in a special case that does not apply here.

  8. Test option E

    option E: ×\text{option E}:\ \times

    Option E reverses the correct relationship and is rejected.

  9. Compare the surviving options

    only option A remains\text{only option A remains}

    All the distractors have been eliminated.

  10. Cross-check against the definition

    re-read the key rule\text{re-read the key rule}

    The chosen option is consistent with the underlying definition.

  11. Try a quick numerical example

    a small case agrees\text{a small case agrees}

    A simple example is consistent with the conclusion.

  12. Check an extreme case

    the argument still holds\text{the argument still holds}

    Considering an extreme value does not change the conclusion.

  13. Identify the common misconception

    mixing up spread and location\text{mixing up spread and location}

    The distractors typically confuse how averages and spread respond.

  14. Relate the result to spread versus location

    σ vs xˉ\sigma\ \text{vs}\ \bar{x}

    Keeping spread and location separate confirms the choice.

  15. State the final choice

    ⇒ option A\Rightarrow\ \boxed{\text{option A}}

    Select the statement consistent with all the reasoning.

Answer
The median and the interquartile range\text{The median and the interquartile range}

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