NCEA Level 2Comparing distributionsAS 91267

Comparing distributions: 7 exam questions solved step by step

What this lesson covers

  • A distribution is described by its shape, centre and spread, each with numbers.
  • Whether a normal model is appropriate depends on the shape of the data.
Key idea

7 questions, one method

  • 1. Shape: symmetric or skewed?
  • 2. Centre: mean or median, with a number.
  • 3. Spread: IQR or sd, with a number.
  • 4. Both groups, in context, every time.
Your turn

Question 1 · Achieved

  • Waiting times at a medical clinic.
  • Describe the shape.
  • Which class holds the median?
  • Which centre should you use?
Question 1 · Achieved
Worked example

Q1: one histogram

  • The tail names the skew.
  • Count to the middle.
  1. 12 + 15 + 10 + 6 + 4 + 2 + 1
  2. = 50 patients
  3. peak: the 10 to 20 min class
  4. long tail to the rightout to 70 min
  5. shape: right-skewed
  6. median: 25th and 26th values
  7. cumulative 12, 2712 < 25, 27 ≥ 26
  8. median in the 10 to 20 min class
  9. skewed: use the median
Your turn

Question 2 · Achieved, in context

  • Commute times, 30 days by bus
  • and 30 days by car.
  • Compare the typical commute times.
BusCar
Minimum38 min30 min
Lower quartile42 min36 min
Median45 min43 min
Upper quartile48 min55 min
Maximum56 min72 min
Worked example

Q2: compare the centres

  • Typical means the centre.
  • Name both, with units.
  1. bus median = 45 minfrom the table
  2. car median = 43 min
  3. 45 − 43 = 2 minthe difference
  4. car median is 2 min lower
Your turn

Question 3 · Merit, in context

  • The same bus and car times.
  • Compare the spreads, with numbers.
  • Which way is more reliable?
BusCar
Minimum38 min30 min
Lower quartile42 min36 min
Median45 min43 min
Upper quartile48 min55 min
Maximum56 min72 min
Worked example

Q3: compare the spreads

  • IQR: the middle half.
  • Spread, not centre.
  1. IQR = UQ − LQ
  2. bus: 48 − 42 = 6 min
  3. car: 55 − 36 = 19 min
  4. range: bus 56 − 38 = 18 min
  5. range: car 72 − 30 = 42 min
  6. 19 ÷ 6 ≈ 3.2times as wide
  7. car times far more variable
  8. the bus is more reliableonly 2 min slower
Your turn

Question 4 · Merit

  • Fish caught on 30 charter trips.
  • Find the mean and the median.
  • What do they say about the shape?
Fish caughtTrips
03
16
28
35
44
52
62
Worked example

Q4: mean against median

  • Mean: Σfx ÷ n.
  • Median: count to the middle.
  1. n = 3 + 6 + 8 + 5 + 4 + 2 + 2 = 30
  2. Σfx = 3×0 + 6×1 + 8×2 + 5×3
  3. + 4×4 + 2×5 + 2×6
  4. = 75
  5. mean = 75 ÷ 30 = 2.5 fish
  6. median: 15th and 16th values
  7. cumulative 3, 9, 179 < 15, 17 ≥ 16
  8. median = 2 fish
  9. mean 2.5 > median 2right-skewed
Your turn

Question 5 · Merit, in context

  • Kiwifruit weights from two orchards.
  • Compare the shape, centre and spread,
  • with numbers.
Orchard AOrchard B
Mean104 g112 g
Median104 g115 g
Standard deviation6 g11 g
Worked example

Q5: shape, centre, spread

  • Mean against median: shape.
  • Skewed: compare medians.
  1. A: mean 104 = median 104symmetric
  2. B: mean 112 < median 115left-skewed
  3. medians: 104 g and 115 gB is skewed
  4. 115 − 104 = 11 g
  5. sd: 6 g and 11 g
  6. 11 ÷ 6 ≈ 1.8nearly twice
  7. Shape: A symmetric, B left-skewed
  8. Centre: B 11 g heaviermedians
  9. Spread: B more variablesd 11 g against 6 g
Your turn

Question 6 · Excellence

  • Back to the clinic waiting times.
  • Is a normal model appropriate?
  • Justify with numbers.
Question 6 · Excellence
Worked example

Q6: the mean and sd

  • Grouped data: midpoints.
  • sd from the calculator.
  1. midpoints 5, 15, 25, …, 65class middles
  2. Σfx = 12×5 + 15×15 + 10×25
  3. + 6×35 + 4×45 + 2×55 + 1×65
  4. = 1100
  5. mean = 1100 ÷ 50 = 22.0 min
  6. s = 15.2864 mincalculator
  7. mean 22.0 > median classright skew
Worked example

Q6: test the model

  • What does the model predict?
  • Compare with reality.
  1. μ = 22.0, σ = 15.2864from the data
  2. z = (0 − 22.0) ÷ 15.2864
  3. = −1.4392
  4. P(X < 0) = 0.0750waits cannot be < 0
  5. model: P(X > 50) = 0.0335
  6. data: 3 ÷ 50 = 0.06the top two classes
  7. normal model not appropriate
Your turn

Question 7 · Excellence

  • Weekly rents ($) for 9 Hamilton flats:
  • 420 450 460 480 490 500 520 540 560
  • Add a tenth flat at $1500 a week.
  • Compare mean and median. Which is better?
Worked example

Q7: before and after

  • Recalculate both.
  • n goes from 9 to 10.
  1. sum = 44209 rents
  2. mean = 4420 ÷ 9 = $491.11
  3. median = 5th value = $490
  4. with $1500: sum = 5920
  5. mean = 5920 ÷ 10 = $592
  6. median = (490 + 500) ÷ 25th and 6th
  7. = $495
Worked example

Q7: which centre?

  • One value, two effects.
  • Say it in context.
  1. mean: 592 − 491.11 = $100.89
  2. median: 495 − 490 = $5
  3. one rent moved the meanthe median resists
  4. 9 of 10 rents are under $592
  5. use the median: $495 a week
Summary

The method that solved all seven

  • Shape, centre and spread: numbers, both groups, in context.
StepHow
Shapesymmetric or skewed, which way
Centrethe median when skewed
SpreadIQR or sd, with numbers
Compareboth groups in every sentence
Normal fitsymmetric, one peak, sensible
Outliersmove the mean, not the median

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