Hard A-Level Large data set fluency Questions

Challenging, exam-style A-Level Large data set fluency questions with worked solutions. Stretch yourself on the hardest standard deviation, iqr, quartiles, range problems.

standard deviationiqrquartilesrangecomparisonmean
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A relative humidity column (%) reads: 70, 72, 71, 73, 72, 70, 71, 99, 72, 71. Using the 1.5×IQR rule, identify the anomalous reading.
Show worked solution

Worked solution

  1. State the outlier rule

    use the 1.5×IQR rule\text{use the }1.5\times\text{IQR rule}

    A value more than 1.5 IQRs beyond a quartile is treated as an anomaly.

  2. List the readings

    data: 70, 72, 71, 73, 72, 70, 71, 99, 72, 71\text{data: }70,\ 72,\ 71,\ 73,\ 72,\ 70,\ 71,\ 99,\ 72,\ 71

    Read the extract as given.

  3. Order the data

    ordered: 70, 70, 71, 71, 71, 72, 72, 72, 73, 99\text{ordered: }70,\ 70,\ 71,\ 71,\ 71,\ 72,\ 72,\ 72,\ 73,\ 99

    Arrange in ascending order.

  4. Count the values

    n=10n=10

    There are 10 readings.

  5. Locate the middle value

    Q2=71.5Q_2=71.5

    With the data ordered, the median is the central value.

  6. Find the lower quartile

    Q1=71Q_1=71

    Q_1 lies one quarter of the way through the ordered data.

  7. Find the upper quartile

    Q3=72Q_3=72

    Q_3 lies three quarters of the way through the ordered data.

  8. Compute the interquartile range

    IQR=Q3−Q1=72−71=1\text{IQR}=Q_3-Q_1=72-71=1

    The IQR measures the spread of the middle 50 percent of the data.

  9. Find the lower fence

    Q1−1.5×IQR=71−1.5×1=69.5Q_1-1.5\times\text{IQR}=71-1.5\times1=69.5

    Anything below this fence is an anomaly.

  10. Find the upper fence

    Q3+1.5×IQR=72+1.5×1=73.5Q_3+1.5\times\text{IQR}=72+1.5\times1=73.5

    Anything above this fence is an anomaly.

  11. Test the smallest reading

    70 vs lower fence 69.570\ \text{vs lower fence }69.5

    Compare the minimum to the lower fence.

  12. Test the largest reading

    99 vs upper fence 73.599\ \text{vs upper fence }73.5

    Compare the maximum to the upper fence.

  13. Identify the value beyond a fence

    anomaly=99\text{anomaly}=99

    This reading falls outside the fences.

  14. Confirm the rest lie in range

    others within [69.5, 73.5]\text{others within }[69.5,\ 73.5]

    All other readings pass the test.

  15. State the anomalous reading

    answer=99\text{answer}=99

    Hence the anomaly is this value.

Answer
99 percent\text{99 percent}
Question 2
8 markschallenging
A windspeed column (knots) reads: 8, 9, 10, 7, 9, 8, 30, 9, 8, 10. Using the 1.5×IQR rule, identify the anomalous reading.
Show worked solution

Worked solution

  1. State the outlier rule

    use the 1.5×IQR rule\text{use the }1.5\times\text{IQR rule}

    A value more than 1.5 IQRs beyond a quartile is treated as an anomaly.

  2. List the readings

    data: 8, 9, 10, 7, 9, 8, 30, 9, 8, 10\text{data: }8,\ 9,\ 10,\ 7,\ 9,\ 8,\ 30,\ 9,\ 8,\ 10

    Read the extract as given.

  3. Order the data

    ordered: 7, 8, 8, 8, 9, 9, 9, 10, 10, 30\text{ordered: }7,\ 8,\ 8,\ 8,\ 9,\ 9,\ 9,\ 10,\ 10,\ 30

    Arrange in ascending order.

  4. Count the values

    n=10n=10

    There are 10 readings.

  5. Locate the middle value

    Q2=9Q_2=9

    With the data ordered, the median is the central value.

  6. Find the lower quartile

    Q1=8Q_1=8

    Q_1 lies one quarter of the way through the ordered data.

  7. Find the upper quartile

    Q3=10Q_3=10

    Q_3 lies three quarters of the way through the ordered data.

  8. Compute the interquartile range

    IQR=Q3−Q1=10−8=2\text{IQR}=Q_3-Q_1=10-8=2

    The IQR measures the spread of the middle 50 percent of the data.

  9. Find the lower fence

    Q1−1.5×IQR=8−1.5×2=5Q_1-1.5\times\text{IQR}=8-1.5\times2=5

    Anything below this fence is an anomaly.

  10. Find the upper fence

    Q3+1.5×IQR=10+1.5×2=13Q_3+1.5\times\text{IQR}=10+1.5\times2=13

    Anything above this fence is an anomaly.

  11. Test the smallest reading

    7 vs lower fence 57\ \text{vs lower fence }5

    Compare the minimum to the lower fence.

  12. Test the largest reading

    30 vs upper fence 1330\ \text{vs upper fence }13

    Compare the maximum to the upper fence.

  13. Identify the value beyond a fence

    anomaly=30\text{anomaly}=30

    This reading falls outside the fences.

  14. Confirm the rest lie in range

    others within [5, 13]\text{others within }[5,\ 13]

    All other readings pass the test.

  15. State the anomalous reading

    answer=30\text{answer}=30

    Hence the anomaly is this value.

Answer
30 knots\text{30 knots}
Question 3
8 markschallenging
A temperature column (°C) reads: 21, 20, 22, 23, 21, 20, 22, 4, 21, 22. Using the 1.5×IQR rule, identify the anomalous reading.
Show worked solution

Worked solution

  1. State the outlier rule

    use the 1.5×IQR rule\text{use the }1.5\times\text{IQR rule}

    A value more than 1.5 IQRs beyond a quartile is treated as an anomaly.

  2. List the readings

    data: 21, 20, 22, 23, 21, 20, 22, 4, 21, 22\text{data: }21,\ 20,\ 22,\ 23,\ 21,\ 20,\ 22,\ 4,\ 21,\ 22

    Read the extract as given.

  3. Order the data

    ordered: 4, 20, 20, 21, 21, 21, 22, 22, 22, 23\text{ordered: }4,\ 20,\ 20,\ 21,\ 21,\ 21,\ 22,\ 22,\ 22,\ 23

    Arrange in ascending order.

  4. Count the values

    n=10n=10

    There are 10 readings.

  5. Locate the middle value

    Q2=21Q_2=21

    With the data ordered, the median is the central value.

  6. Find the lower quartile

    Q1=20Q_1=20

    Q_1 lies one quarter of the way through the ordered data.

  7. Find the upper quartile

    Q3=22Q_3=22

    Q_3 lies three quarters of the way through the ordered data.

  8. Compute the interquartile range

    IQR=Q3−Q1=22−20=2\text{IQR}=Q_3-Q_1=22-20=2

    The IQR measures the spread of the middle 50 percent of the data.

  9. Find the lower fence

    Q1−1.5×IQR=20−1.5×2=17Q_1-1.5\times\text{IQR}=20-1.5\times2=17

    Anything below this fence is an anomaly.

  10. Find the upper fence

    Q3+1.5×IQR=22+1.5×2=25Q_3+1.5\times\text{IQR}=22+1.5\times2=25

    Anything above this fence is an anomaly.

  11. Test the smallest reading

    4 vs lower fence 174\ \text{vs lower fence }17

    Compare the minimum to the lower fence.

  12. Test the largest reading

    23 vs upper fence 2523\ \text{vs upper fence }25

    Compare the maximum to the upper fence.

  13. Identify the value beyond a fence

    anomaly=4\text{anomaly}=4

    This reading falls outside the fences.

  14. Confirm the rest lie in range

    others within [17, 25]\text{others within }[17,\ 25]

    All other readings pass the test.

  15. State the anomalous reading

    answer=4\text{answer}=4

    Hence the anomaly is this value.

Answer
4 °C\text{4 °C}
Question 4
8 markschallenging
A day's rainfall column (mm) reads: 1, 2, 0, 3, 2, 1, 2, 3, 1, 25. Using the 1.5×IQR rule, identify the anomalous reading.
Show worked solution

Worked solution

  1. State the outlier rule

    use the 1.5×IQR rule\text{use the }1.5\times\text{IQR rule}

    A value more than 1.5 IQRs beyond a quartile is treated as an anomaly.

  2. List the readings

    data: 1, 2, 0, 3, 2, 1, 2, 3, 1, 25\text{data: }1,\ 2,\ 0,\ 3,\ 2,\ 1,\ 2,\ 3,\ 1,\ 25

    Read the extract as given.

  3. Order the data

    ordered: 0, 1, 1, 1, 2, 2, 2, 3, 3, 25\text{ordered: }0,\ 1,\ 1,\ 1,\ 2,\ 2,\ 2,\ 3,\ 3,\ 25

    Arrange in ascending order.

  4. Count the values

    n=10n=10

    There are 10 readings.

  5. Locate the middle value

    Q2=2Q_2=2

    With the data ordered, the median is the central value.

  6. Find the lower quartile

    Q1=1Q_1=1

    Q_1 lies one quarter of the way through the ordered data.

  7. Find the upper quartile

    Q3=3Q_3=3

    Q_3 lies three quarters of the way through the ordered data.

  8. Compute the interquartile range

    IQR=Q3−Q1=3−1=2\text{IQR}=Q_3-Q_1=3-1=2

    The IQR measures the spread of the middle 50 percent of the data.

  9. Find the lower fence

    Q1−1.5×IQR=1−1.5×2=−2Q_1-1.5\times\text{IQR}=1-1.5\times2=-2

    Anything below this fence is an anomaly.

  10. Find the upper fence

    Q3+1.5×IQR=3+1.5×2=6Q_3+1.5\times\text{IQR}=3+1.5\times2=6

    Anything above this fence is an anomaly.

  11. Test the smallest reading

    0 vs lower fence −20\ \text{vs lower fence }-2

    Compare the minimum to the lower fence.

  12. Test the largest reading

    25 vs upper fence 625\ \text{vs upper fence }6

    Compare the maximum to the upper fence.

  13. Identify the value beyond a fence

    anomaly=25\text{anomaly}=25

    This reading falls outside the fences.

  14. Confirm the rest lie in range

    others within [−2, 6]\text{others within }[-2,\ 6]

    All other readings pass the test.

  15. State the anomalous reading

    answer=25\text{answer}=25

    Hence the anomaly is this value.

Answer
25 mm\text{25 mm}
Question 5
8 markschallenging
Location A recorded daily mean temperatures (°C) of 22, 24, 26, 28, 24, 26, 22, 28 and Location B recorded 15, 17, 19, 21, 17, 19, 15, 21 over the same 8 days. After summarising both locations, find the difference between their mean temperatures (A minus B).
Show worked solution

Worked solution

  1. Set up the comparison

    compare means of A and B\text{compare means of A and B}

    We compute a summary for each location then compare.

  2. Total location A

    ∑xA=22+22+24+24+26+26+28+28=200\sum x_A=22+22+24+24+26+26+28+28=200

    Add the readings for A.

  3. Mean of A

    xˉA=2008=25\bar{x}_A=\dfrac{200}{8}=25

    Divide A's total by its count.

  4. Order location A

    A ordered: 22, 22, 24, 24, 26, 26, 28, 28\text{A ordered: }22,\ 22,\ 24,\ 24,\ 26,\ 26,\ 28,\ 28

    Sort A to read off spread.

  5. Range of A

    rangeA=28−22=6\text{range}_A=28-22=6

    Largest minus smallest for A.

  6. IQR of A

    IQRA=27−23=4\text{IQR}_A=27-23=4

    Spread of the middle 50 percent of A.

  7. Total location B

    ∑xB=15+15+17+17+19+19+21+21=144\sum x_B=15+15+17+17+19+19+21+21=144

    Add the readings for B.

  8. Mean of B

    xˉB=1448=18\bar{x}_B=\dfrac{144}{8}=18

    Divide B's total by its count.

  9. Order location B

    B ordered: 15, 15, 17, 17, 19, 19, 21, 21\text{B ordered: }15,\ 15,\ 17,\ 17,\ 19,\ 19,\ 21,\ 21

    Sort B to read off spread.

  10. Range of B

    rangeB=21−15=6\text{range}_B=21-15=6

    Largest minus smallest for B.

  11. IQR of B

    IQRB=20−16=4\text{IQR}_B=20-16=4

    Spread of the middle 50 percent of B.

  12. Compare the spreads

    IQRA=4, IQRB=4\text{IQR}_A=4,\ \text{IQR}_B=4

    State which location is more variable by IQR.

  13. Standard deviation of A

    σA=2.24\sigma_A=2.24

    A's spread measured by standard deviation.

  14. Standard deviation of B

    σB=2.24\sigma_B=2.24

    B's spread measured by standard deviation.

  15. Difference in means

    xˉA−xˉB=25−18=7\bar{x}_A-\bar{x}_B=25-18=7

    The required difference between the two location means.

Answer
77

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