Hard A-Level Measures of location and spread Questions

Challenging, exam-style A-Level Measures of location and spread questions with worked solutions. Stretch yourself on the hardest mean, standard deviation, grouped data, median problems.

meanstandard deviationgrouped datamedianinterpolationquartiles
A-Level34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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}
Question 2
8 markschallenging
Machine P produces components with mean length 5050 mm and standard deviation 0.20.2 mm; Machine Q has mean 5050 mm and standard deviation 0.50.5 mm. Which machine is more consistent?
Show worked solution

Worked solution

  1. Compare the means

    xˉP=xˉQ=50\bar{x}_P = \bar{x}_Q = 50

    Both machines have the same average length.

  2. Compare the standard deviations

    σP=0.2<σQ=0.5\sigma_P = 0.2 < \sigma_Q = 0.5

    A smaller standard deviation means less variation.

  3. Interpret consistency

    ⇒ Machine P more consistent\Rightarrow\ \text{Machine P more consistent}

    Less spread about the mean means more consistent output.

  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
Machine P, because its standard deviation is smaller\text{Machine P, because its standard deviation is smaller}
Question 3
8 markschallenging
Data are coded using y=x−1004y = \dfrac{x-100}{4}. The variance of yy is 6.256.25. What is the variance of xx?
Show worked solution

Worked solution

  1. Rearrange the coding

    x=4y+100x = 4y + 100

    Make xx the subject.

  2. Effect of coding on variance

    σx2=42 σy2\sigma_x^2 = 4^2\,\sigma_y^2

    Variance scales by the square of the multiplier; the shift has no effect.

  3. Substitute the numbers

    σx2=16×6.25=100\sigma_x^2 = 16\times 6.25 = 100

    The variance of xx is 100100.

  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
100\text{100}
Question 4
8 markschallenging
Class AA: 5050 values, mean 2020, standard deviation 66. Class BB: 5050 values, mean 3030, standard deviation 66. Find the combined mean and standard deviation.
Show worked solution

Worked solution

  1. Find ∑x\sum x for sample A

    ∑xA=nAxˉA=50×20=1000\sum x_A = n_A\bar{x}_A = 50\times 20 = 1000

    Use ∑x=nxˉ\sum x = n\bar{x}.

  2. Find ∑x2\sum x^2 for sample A

    ∑xA2=nA(σA2+xˉA2)=50(62+202)=21800\sum x_A^2 = n_A(\sigma_A^2+\bar{x}_A^2) = 50(6^2+20^2) = 21800

    Rearrange σ2=∑x2/n−xˉ2\sigma^2 = \sum x^2/n - \bar{x}^2.

  3. Find ∑x\sum x for sample B

    ∑xB=50×30=1500\sum x_B = 50\times 30 = 1500

    Same method for sample B.

  4. Find ∑x2\sum x^2 for sample B

    ∑xB2=50(62+302)=46800\sum x_B^2 = 50(6^2+30^2) = 46800

    Same method for sample B.

  5. Find the combined count

    n=50+50=100n = 50+50 = 100

    Add the sample sizes.

  6. Find the combined ∑x\sum x

    ∑x=1000+1500=2500\sum x = 1000+1500 = 2500

    Add the two totals.

  7. Find the combined mean

    xˉ=2500100=25\bar{x} = \frac{2500}{100} = 25

    Divide the combined total by nn.

  8. Find the combined ∑x2\sum x^2

    ∑x2=21800+46800=68600\sum x^2 = 21800+46800 = 68600

    Add the two sums of squares.

  9. Find the combined SxxS_{xx}

    Sxx=68600−25002100=6100S_{xx} = 68600 - \frac{2500^2}{100} = 6100

    Use Sxx=∑x2−(∑x)2/nS_{xx}=\sum x^2-(\sum x)^2/n.

  10. Find the combined variance

    σ2=6100100=61\sigma^2 = \frac{6100}{100} = 61

    Divide SxxS_{xx} by nn.

  11. Find the combined standard deviation

    σ=61=7.81\sigma = \sqrt{61} = 7.81

    Square root the variance.

  12. Check the mean is a weighted average

    50×20+50×30100=25\frac{50\times20+50\times30}{100} = 25

    The combined mean lies between the two group means.

  13. Comment on the combined spread

    σ reflects both within- and between-group variation\sigma\text{ reflects both within- and between-group variation}

    Differing group means add to the overall spread.

  14. State the combined mean

    xˉ=25\bar{x} = 25

    Final combined mean.

  15. State the combined standard deviation

    σ=7.81\sigma = 7.81

    Final combined standard deviation.

Answer
xˉ=25, σ=7.81\bar{x}=25,\ \sigma=7.81
Question 5
8 markschallenging
Set AA: 2525 values, mean 88, standard deviation 22. Set BB: 1515 values, mean 1212, standard deviation 33. Find the combined mean and standard deviation.
Show worked solution

Worked solution

  1. Find ∑x\sum x for sample A

    ∑xA=nAxˉA=25×8=200\sum x_A = n_A\bar{x}_A = 25\times 8 = 200

    Use ∑x=nxˉ\sum x = n\bar{x}.

  2. Find ∑x2\sum x^2 for sample A

    ∑xA2=nA(σA2+xˉA2)=25(22+82)=1700\sum x_A^2 = n_A(\sigma_A^2+\bar{x}_A^2) = 25(2^2+8^2) = 1700

    Rearrange σ2=∑x2/n−xˉ2\sigma^2 = \sum x^2/n - \bar{x}^2.

  3. Find ∑x\sum x for sample B

    ∑xB=15×12=180\sum x_B = 15\times 12 = 180

    Same method for sample B.

  4. Find ∑x2\sum x^2 for sample B

    ∑xB2=15(32+122)=2295\sum x_B^2 = 15(3^2+12^2) = 2295

    Same method for sample B.

  5. Find the combined count

    n=25+15=40n = 25+15 = 40

    Add the sample sizes.

  6. Find the combined ∑x\sum x

    ∑x=200+180=380\sum x = 200+180 = 380

    Add the two totals.

  7. Find the combined mean

    xˉ=38040=9.5\bar{x} = \frac{380}{40} = 9.5

    Divide the combined total by nn.

  8. Find the combined ∑x2\sum x^2

    ∑x2=1700+2295=3995\sum x^2 = 1700+2295 = 3995

    Add the two sums of squares.

  9. Find the combined SxxS_{xx}

    Sxx=3995−380240=385S_{xx} = 3995 - \frac{380^2}{40} = 385

    Use Sxx=∑x2−(∑x)2/nS_{xx}=\sum x^2-(\sum x)^2/n.

  10. Find the combined variance

    σ2=38540=9.62\sigma^2 = \frac{385}{40} = 9.62

    Divide SxxS_{xx} by nn.

  11. Find the combined standard deviation

    σ=9.62=3.1\sigma = \sqrt{9.62} = 3.1

    Square root the variance.

  12. Check the mean is a weighted average

    25×8+15×1240=9.5\frac{25\times8+15\times12}{40} = 9.5

    The combined mean lies between the two group means.

  13. Comment on the combined spread

    σ reflects both within- and between-group variation\sigma\text{ reflects both within- and between-group variation}

    Differing group means add to the overall spread.

  14. State the combined mean

    xˉ=9.5\bar{x} = 9.5

    Final combined mean.

  15. State the combined standard deviation

    σ=3.1\sigma = 3.1

    Final combined standard deviation.

Answer
xˉ=9.5, σ=3.1\bar{x}=9.5,\ \sigma=3.1

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