A-Level Large data set fluency Practice Questions

Free A-Level Large data set fluency practice questions with full step-by-step worked solutions. Covers mean, summary statistics, median, range. Practise exam-style problems and check your method.

meansummary statisticsmedianrangespreadmaximum
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
The daily mean temperature (°C) recorded at Heathrow on 8 days was: 15, 17, 16, 18, 14, 16, 15, 17. Calculate the mean daily temperature.
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Worked solution

  1. Write the mean formula

    xˉ=∑xn\bar{x}=\dfrac{\sum x}{n}

    The mean equals the total of the readings divided by how many there are.

  2. Add all the readings

    ∑x=14+15+15+16+16+17+17+18=128\sum x=14+15+15+16+16+17+17+18=128

    Total the readings in the extract.

  3. Divide the total by n

    xˉ=1288=16\bar{x}=\dfrac{128}{8}=16

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

Answer
1616
Question 2
2 markseasy
In the rainfall column the entry 'tr' is used. What does it mean?
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Worked solution

  1. Interpret the question

    code tr\text{code tr}

    'tr' is a special rainfall code.

  2. Recall the relevant idea

    trace amount\text{trace amount}

    It means a trace of rain was detected.

  3. State the correct choice

    correct: A trace: rainfall too small to measure (less than 0.05 mm)\text{correct: A trace: rainfall too small to measure (less than 0.05 mm)}

    Therefore the correct answer is: A trace: rainfall too small to measure (less than 0.05 mm).

Answer
A trace: rainfall too small to measure (less than 0.05 mm)\text{A trace: rainfall too small to measure (less than 0.05 mm)}
Question 3
3 marksintermediate
State a limitation of relying on the large data set.
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Worked solution

  1. Interpret the question

    selected stations\text{selected stations}

    Only certain weather stations are included.

  2. Recall the relevant idea

    selected periods\text{selected periods}

    Only particular time periods are covered.

  3. Apply it to this data

    may not generalise\text{may not generalise}

    So conclusions may not apply everywhere.

  4. Rule out option (B)

    reject: It contains every possible weather value\text{reject: It contains every possible weather value}

    Option (B) (It contains every possible weather value) does not satisfy the definition, so it is rejected.

  5. Rule out option (C)

    reject: It has no measurement units\text{reject: It has no measurement units}

    Option (C) (It has no measurement units) does not satisfy the definition, so it is rejected.

  6. State the correct choice

    correct: It only covers selected locations and time periods, so may not generalise\text{correct: It only covers selected locations and time periods, so may not generalise}

    Therefore the correct answer is: It only covers selected locations and time periods, so may not generalise.

Answer
It only covers selected locations and time periods, so may not generalise\text{It only covers selected locations and time periods, so may not generalise}
Question 4
5 markshard
Why might a sample of only summer days misrepresent the whole data set?
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Worked solution

  1. Interpret the question

    summer only\text{summer only}

    The sample covers a single season.

  2. Recall the relevant idea

    ignores other seasons\text{ignores other seasons}

    Weather varies across the year.

  3. Apply it to this data

    seasonal bias\text{seasonal bias}

    So it misrepresents the full period.

  4. Rule out option (B)

    reject: Because summer has no data\text{reject: Because summer has no data}

    Option (B) (Because summer has no data) does not satisfy the definition, so it is rejected.

  5. Rule out option (C)

    reject: Because summer data is qualitative\text{reject: Because summer data is qualitative}

    Option (C) (Because summer data is qualitative) does not satisfy the definition, so it is rejected.

  6. Rule out option (D)

    reject: Because it doubles the mean\text{reject: Because it doubles the mean}

    Option (D) (Because it doubles the mean) does not satisfy the definition, so it is rejected.

  7. Rule out option (E)

    reject: Because it removes units\text{reject: Because it removes units}

    Option (E) (Because it removes units) does not satisfy the definition, so it is rejected.

  8. Check the units are appropriate

    units consistent with the quantity\text{units consistent with the quantity}

    The chosen answer must use units matching what is being recorded.

  9. Confirm the data type

    counted vs measured\text{counted vs measured}

    Decide whether the values are counted (discrete) or measured (continuous).

  10. State the correct choice

    correct: Because it ignores seasonal variation across the whole period\text{correct: Because it ignores seasonal variation across the whole period}

    Therefore the correct answer is: Because it ignores seasonal variation across the whole period.

Answer
Because it ignores seasonal variation across the whole period\text{Because it ignores seasonal variation across the whole period}
Question 5
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.
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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}

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