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Cumulative original practice combining sample design, grouped estimates, percentile interpolation, coding and careful interpretation of weather summaries.
Before you startData collection, measures of location and spread, and data representations.
01 / Identify the assumption behind each method
Some answers are exact descriptions; others are estimates.
Distinguish proportional sample calculations from assumptions about within-class positions or the relevance of a dataset. The numerical datasets here are constructed for practice.
A constructed organisation has 400 technical and 800 other staff. A survey needs 60 staff opinions.
02 / Choose a sample that fits the population
Explain one benefit and one cost of surveying all 1200 staff.
A census includes the full target population.
It avoids selecting only a subset, but collecting and processing every response takes more time and resources. Nonresponse or inaccurate answers can still affect a census.
Find the numbers in a proportionally stratified sample of 60 from groups of 400 and 800.
Multiply 60 by each population fraction.
Technical: 60×400/1200=20. Other staff: 60×800/1200=40. Select appropriately within each stratum.
Why might the first 60 arrivals misrepresent all staff?
Arrival time may relate to job role or shift.
It favours particular arrival patterns and may exclude later shifts. Increasing this convenient sample does not automatically remove the selection bias.
Describe a systematic sample of 60 from an ordered list of 1200 names.
Use an interval and a random start.
Choose a random start from 1 to 20, then take every twentieth name. Check that the list ordering has no periodic pattern aligned with the interval.
03 / Combine grouped means and standard deviations
A constructed table has classes 0≤t<10, 10≤t<20 and 20≤t<40, with frequencies 4, 8 and 8. Use midpoints 5, 15 and 30. Here n=20, Σft=380 and Σft²=9100, where t denotes the class midpoint in these summaries.
Estimate the mean of t.
Divide the midpoint-weighted total by the frequency.
380/20=19. It is an estimate because exact within-class values are unknown.
Estimate the standard deviation using division by n.
Variance = Σft²/n − mean².
Variance = 9100/20−19²=94. Standard deviation = √94≈9.695. Its units match those of t.
Why would using 25 for the final class midpoint be wrong?
Find the midpoint of both endpoints.
The final class is 20 to 40, so its midpoint is (20+40)/2=30. Unequal widths must be respected.
Why is Σft² not equal to (Σft)²?
One sums weighted squares; the other squares a whole sum.
Σft²=9100 sums each midpoint square times its frequency. (Σft)²=380² is a different quantity and cannot replace it in the variance formula.
04 / Interpolate using cumulative position
Pause, replay or seek freely. The notes explain the same idea and stay in view.
Median position = 20/2=10
This lies in the 10 to 20 class.
Median ≈ 10 + ((10−4)/8)×10 = 17.5
The fraction of this class needed is 6/8.
20th percentile position = 4; 80th = 16
P₂₀≈10 and P₈₀≈20+((16−12)/8)×20=30. The interpercentile range is about 20.
What assumption supports these percentile estimates?
How are observations treated within a class?
They are assumed spread uniformly within each class. Raw values could produce different percentiles.
Why might calculation from the table be more precise than reading a hand-drawn cumulative-frequency graph?
Both methods estimate, but one adds drawing and reading error.
The calculation avoids extra plotting and visual-reading error while retaining the same within-class approximation.
05 / Undo a transformation in the correct order
A constructed dataset uses u=2t+3. For 10 observations, Σu=230 and Σu²=5590. Use population-style variance with division by n.
Calculate the mean and variance of u.
Use the totals before reversing the coding.
Mean u=230/10=23. Variance u=5590/10−23²=30.
Recover the mean and standard deviation of t.
t=(u−3)/2.
Mean t=(23−3)/2=10. Standard deviation t=√30/2=√7.5≈2.739. Do not subtract 3 from the standard deviation.
06 / A summary alone does not justify a broad weather claim
A learner takes the first five recorded days of a month and generalises to the whole season. Name two limitations.
Consider selection and dependence.
The dates may not represent the season; nearby days may share weather conditions. The short sample and any missing observations also limit inference.
Why does a mean of daily maximum humidity not directly give the proportion of misty days?
The summary and the requested event are different.
A mean of maxima does not count mist events or retain their timing. A justified event definition and appropriate records would be needed.
07 / Distinguish arithmetic errors from modelling limits
If t measures minutes, what units do its variance and standard deviation have?
Squaring changes units.
Variance is in minutes squared; standard deviation is in minutes.
Which is exact from the stated table: total frequency 20, grouped mean 19, or estimated median 17.5?
Only one requires no assumption about within-class values.
Total frequency 20 is exact. The grouped mean and interpolated median estimate summaries of the unknown raw observations.
08 / Choose methods deliberately
Sampling questions need a defined population and a defensible selection method. Grouped summaries need the correct midpoints and cumulative frequencies. Coding changes location and scale differently. Interpret weather variables and sample periods before making population claims.
Section 1 of 8 · Identify the assumption behind each method