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Estimation station and class survey for IB Maths

A first lesson that sets the tone for the course: estimate first, then measure, and treat being wrong as part of maths. Then collect some anonymous class data and model it with Pearson’s r and a regression line.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Approximations, bounds and percentage error; Descriptive statistics: mean and standard deviation; Correlation and regression
Equipment
The starter is non-calculator. A GDC is needed for tasks B and C.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A10 minEstimation station
Main: task B15 minClass survey: correlation and regression
Main: task C10 minTravel times
Extension10 minFast finishers or homework

Starter (5 minutes)

Estimate, don’t calculate: round first.

  1. Estimate 3.982 × 25.1 by rounding each number to the nearest whole number.
  2. Estimate √99 × √401.
  3. A student estimates that a corridor is 30 m long. It is really 24 m. Find the percentage error.

Main activity (35 minutes)

Task A: Estimation station (10 min)

At each station, write your estimate first, then measure. These are one group’s results.

  1. Five estimates of the number of counters in a jar were 150, 210, 180, 260 and 200. Find the mean and the standard deviation of the estimates.
  2. There are 196 counters. Find the percentage error of the mean estimate.
  3. A corridor is 24 m long to the nearest metre and 2.5 m wide to the nearest 0.1 m. Find the upper bound of its floor area.

Task B: Class survey: correlation and regression (15 min)

Made-up data from eight students:

Height, h cm150155160165170175180185
Hand span, s cm1718182020212323

  1. Find Pearson’s product-moment correlation coefficient, r, and describe the correlation.
  2. Find the equation of the regression line of s on h.
  3. Use your line to estimate the hand span of a student who is 168 cm tall.
  4. Explain why the line should not be used to estimate the hand span of a child who is 120 cm tall.

Task C: Travel times (10 min)

Made-up data: a class of 25 recorded how long their journey to school takes.

Time, t minutesFrequency
0 < t ≤ 106
10 < t ≤ 209
20 < t ≤ 307
30 < t ≤ 403

  1. Estimate the mean and the standard deviation of the journey times.
  2. Which class interval contains the median?

Extension (10 minutes)

For fast finishers.

  1. Every hand span in task B was measured 0.5 cm too short. Write down the corrected value of r and the corrected regression line.

For teachers

Teacher notes and full worked answers
  • Estimation station: set up four or five stations (a desk to measure, a jar of counters, a corridor, a book’s thickness, a minute with eyes closed). Students write an estimate first, then measure. Being wrong is the point: celebrate the closest estimate and the best reasoning, not just the right answer, and agree a class norm that a reasoned wrong answer is a good start.
  • Class survey: the data in the tasks are made up, so the answers can be checked. Then collect your own class’s anonymous data on the board (a show of hands for travel-time groups, hand spans written on sticky notes with no names) and repeat the calculations. Nothing is stored on the site.
  • For AI students, add Spearman’s rank on the class’s own data; for everyone, ask which variable should be x and why.

Starter

  1. About 400
    • 42 × 25 = 400
  2. About 200
    • 10 × 20 = 200
  3. 25%
    • |30 − 24|/24 × 100 = 25%

Task A: Estimation station

  1. Mean 200; standard deviation 36.3
    • Mean = 1000 ÷ 5 = 200
    • Deviations −50, 10, −20, 60, 0; squares add to 6600
    • σ = √(6600 ÷ 5) = √1320 = 36.3
  2. 2.04%
    • |200 − 196|/196 × 100 = 2.04%
  3. 62.475 m2
    • Upper bounds 24.5 m and 2.55 m
    • 24.5 × 2.55 = 62.475

Task B: Class survey: correlation and regression

  1. r = 0.977: strong positive correlation
    • From the GDC: r = 0.97725…
  2. s = 0.181h − 10.3
    • From the GDC: s = 0.180952…h − 10.3095…
  3. 20.1 cm
    • 0.180952… × 168 − 10.3095… = 20.09…
  4. 120 cm is outside the data (150–185 cm)
    • That would be extrapolation: the linear pattern may not hold outside the data.

Task C: Travel times

  1. Mean 17.8 minutes; standard deviation 9.6 minutes
    • Midpoints 5, 15, 25, 35 with frequencies 6, 9, 7, 3
    • Mean = 445 ÷ 25 = 17.8
    • σ = √(10 225 ÷ 25 − 17.82) = √92.16 = 9.6
  2. 10 < t ≤ 20
    • The median is the 13th value; running totals 6, 15.

Extension

  1. r = 0.977 (unchanged); s = 0.181h − 9.81
    • Adding 0.5 to every s does not change how the points spread, so r is unchanged.
    • The line moves up 0.5: the intercept becomes −10.3095… + 0.5 = −9.81

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

Practise the topics

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