Mathematics · Simulation and virtual lab Free
Statistics explorer: averages, spread and graphs
Show the same data as a dot plot, histogram, box plot or cumulative frequency curve, add your own values and outliers, and watch the mean, median, quartiles and spread change.
- You will learn to
- Represent data with dot plots, histograms, frequency polygons, box plots and cumulative frequency curves.
- Calculate and compare the mean, median, mode, quartiles, range, IQR and standard deviation.
- See how outliers and skewed data affect each average, and describe correlation.
- You will measure
- Mean, median, mode, quartiles, range, IQR and σ for each data set; r for two variables.
- Time needed
- About 30–40 min
- Equipment
- Nothing extra. Real version: tape measure, ruler for the ruler-drop test, class data collected by the learners.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To represent data with dot plots, histograms, frequency polygons, box plots and cumulative frequency curves; to calculate and compare the mean, median, mode, quartiles, range, interquartile range and standard deviation; and to describe the correlation between two variables.
What to do
- Choose Heights of 40 S1 learners and the dot plot. Read the mean, median and mode. Tap Record the statistics.
- Tap Add an outlier and record again. Which changed more, the mean or the median? Remove the outlier by tapping its dot.
- Show the data as a histogram with class widths 3, 5 and 10 cm. Which class width shows the shape best? Find the modal class.
- Show the box plot and the cumulative frequency curve. Read Q₁, the median and Q₃ from the curve and compare with the read-outs.
- Choose Daily rainfall in Kigali. Compare the mean and the median: which is the better “typical” value, and why?
- Choose Ruler-drop catch distances and tap Do a ruler drop five times. Then (S5) choose Height and arm span and describe the correlation.
Results
Questions
Find the mean, median and mode of 4, 7, 7, 9, 13.
Mean = 40 ÷ 5 = 8; median = 7 (the middle value); mode = 7 (the most frequent).
Why does an outlier change the mean much more than the median?
The mean uses the size of every value, so one very large value pulls it up. The median only depends on the middle position, which moves by at most half a place.
Why is the median a better average for the Kigali rainfall data?
The data are skewed: most days have little rain and a few have a lot. The few large values pull the mean above most days, so the median describes a typical day better.
What does the interquartile range tell you, and why is it often better than the range?
IQR = Q₃ − Q₁ is the spread of the middle half of the data. It ignores the extreme values, so one unusual value does not change it much.
How do you find the median from a cumulative frequency curve?
Go across from n ÷ 2 on the cumulative frequency axis to the curve, then down to the horizontal axis and read the value. Use n ÷ 4 and 3n ÷ 4 for the quartiles.
What does r = 0.9 mean for height and arm span?
There is a strong positive correlation: taller learners usually have longer arm spans, and the points lie close to the line of best fit.
For teachers: expected results and how the model works
The data sets are fixed (the same for every learner) so that results can be compared in class: heights n = 40: mean 150.5 cm, median 151.5 cm, mode 155 cm, range 45 cm, IQR 8 cm, σ = 7.81 cm (adding the outlier 180 cm moves the mean by 0.72 cm but the median by only 0.5 cm); marks n = 40: mean 32.2, median 29; rainfall n = 30: mean 4.1 mm but median 1.5 mm (skewed, many dry days); ruler drop n = 24: mean 16.7 cm, median 17 cm (each “Do a ruler drop” adds a random catch from 6 to 30 cm). Quartiles use the median of each half (for odd n the middle value is left out); σ is the population standard deviation √(Σ(x − x̄)² ÷ n). The cumulative frequency curve is drawn with straight lines between upper class boundaries, so values read from it are estimates. Height and arm span: r = 0.953, best-fit line y = −7.2 + 1.05x through (x̄, ȳ) = (151.9, 152.1).
Exam practice Exam set
National-exam style questions on this simulation (19 marks). Try each one on paper first, then open the marking scheme.
The marks of 10 learners are 12, 15, 15, 18, 20, 21, 24, 25, 30, 45. (a) Find the mean, median and mode. (b) Find the quartiles and the interquartile range. (c) Which average best represents the marks? Give a reason. (7 marks)
Marking scheme: (a) Mean = 225 ÷ 10 = 22.5 (1); median = (20 + 21) ÷ 2 = 20.5 (1); mode = 15 (1). (b) Q₁ = 15, Q₃ = 25 (1, 1); IQR = 10 (1). (c) The median, because the outlier 45 pulls the mean up (1).
Calculate the standard deviation of 2, 4, 4, 4, 5, 5, 7, 9. (3 marks)
Marking scheme: Mean = 40 ÷ 8 = 5 (1); Σ(x − x̄)² = 9 + 1 + 1 + 1 + 0 + 0 + 4 + 16 = 32 (1); σ = √(32 ÷ 8) = √4 = 2 (1).
Heights (cm) of 40 learners: 140–145: 3; 145–150: 9; 150–155: 14; 155–160: 10; 160–165: 4. (a) Draw a cumulative frequency table. (b) Estimate the median from a cumulative frequency curve. (c) Estimate the mean using mid-points. (7 marks)
Marking scheme: (a) Upper boundaries 145, 150, 155, 160, 165 with cumulative frequencies 3, 12, 26, 36, 40 (2). (b) Median at the 20th value: 150 + (20 − 12) ÷ 14 × 5 ≈ 152.9 cm (2). (c) Σfx = 3 × 142.5 + 9 × 147.5 + 14 × 152.5 + 10 × 157.5 + 4 × 162.5 = 6 120 (1); mean = 6 120 ÷ 40 = 153 cm (2).
A ruler is caught after falling 18 cm. Calculate the reaction time. (g = 9.8 m/s²) (2 marks)
Marking scheme: d = ½gt², so t = √(2d ÷ g) = √(2 × 0.18 ÷ 9.8) (1) = 0.19 s (1).
Donat Sciences and MathsMathematics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Statistics explorer: averages, spread and graphs
Aim: To represent data with dot plots, histograms, frequency polygons, box plots and cumulative frequency curves; to calculate and compare the mean, median, mode, quartiles, range, interquartile range and standard deviation; and to describe the correlation between two variables.
Apparatus (real laboratory): tape measure, ruler for the ruler-drop test, class data collected by the learners.
What to do (with the simulation on www.donatstem.co.rw)
- Choose Heights of 40 S1 learners and the dot plot. Read the mean, median and mode. Tap Record the statistics.
- Tap Add an outlier and record again. Which changed more, the mean or the median? Remove the outlier by tapping its dot.
- Show the data as a histogram with class widths 3, 5 and 10 cm. Which class width shows the shape best? Find the modal class.
- Show the box plot and the cumulative frequency curve. Read Q₁, the median and Q₃ from the curve and compare with the read-outs.
- Choose Daily rainfall in Kigali. Compare the mean and the median: which is the better “typical” value, and why?
- Choose Ruler-drop catch distances and tap Do a ruler drop five times. Then (S5) choose Height and arm span and describe the correlation.
Results
| # | Data set | n | Mean | Median | Mode | Range | IQR | σ (or r) |
|---|
| 1 | | | | | | | | |
| 2 | | | | | | | | |
| 3 | | | | | | | | |
| 4 | | | | | | | | |
| 5 | | | | | | | | |
| 6 | | | | | | | | |
| 7 | | | | | | | | |
| 8 | | | | | | | | |
Graph
Draw a histogram of the height data with class width 5 cm and add the frequency polygon. On a second grid, draw the cumulative frequency curve and read Q₁, the median and Q₃.
Questions
- Find the mean, median and mode of 4, 7, 7, 9, 13. [2]
- Why does an outlier change the mean much more than the median? [2]
- Why is the median a better average for the Kigali rainfall data? [2]
- What does the interquartile range tell you, and why is it often better than the range? [2]
- How do you find the median from a cumulative frequency curve? [2]
- What does r = 0.9 mean for height and arm span? [2]
- The marks of 10 learners are 12, 15, 15, 18, 20, 21, 24, 25, 30, 45. (a) Find the mean, median and mode. (b) Find the quartiles and the interquartile range. (c) Which average best represents the marks? Give a reason. [7]
- Calculate the standard deviation of 2, 4, 4, 4, 5, 5, 7, 9. [3]
- Heights (cm) of 40 learners: 140–145: 3; 145–150: 9; 150–155: 14; 155–160: 10; 160–165: 4. (a) Draw a cumulative frequency table. (b) Estimate the median from a cumulative frequency curve. (c) Estimate the mean using mid-points. [7]
- A ruler is caught after falling 18 cm. Calculate the reaction time. (g = 9.8 m/s²) [2]
Donat Sciences and MathsMathematics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Statistics explorer: averages, spread and graphs
Expected results
The data sets are fixed (the same for every learner) so that results can be compared in class: heights n = 40: mean 150.5 cm, median 151.5 cm, mode 155 cm, range 45 cm, IQR 8 cm, σ = 7.81 cm (adding the outlier 180 cm moves the mean by 0.72 cm but the median by only 0.5 cm); marks n = 40: mean 32.2, median 29; rainfall n = 30: mean 4.1 mm but median 1.5 mm (skewed, many dry days); ruler drop n = 24: mean 16.7 cm, median 17 cm (each “Do a ruler drop” adds a random catch from 6 to 30 cm). Quartiles use the median of each half (for odd n the middle value is left out); σ is the population standard deviation √(Σ(x − x̄)² ÷ n). The cumulative frequency curve is drawn with straight lines between upper class boundaries, so values read from it are estimates. Height and arm span: r = 0.953, best-fit line y = −7.2 + 1.05x through (x̄, ȳ) = (151.9, 152.1).
Answers
- Find the mean, median and mode of 4, 7, 7, 9, 13. [2]
Mean = 40 ÷ 5 = 8; median = 7 (the middle value); mode = 7 (the most frequent).
- Why does an outlier change the mean much more than the median? [2]
The mean uses the size of every value, so one very large value pulls it up. The median only depends on the middle position, which moves by at most half a place.
- Why is the median a better average for the Kigali rainfall data? [2]
The data are skewed: most days have little rain and a few have a lot. The few large values pull the mean above most days, so the median describes a typical day better.
- What does the interquartile range tell you, and why is it often better than the range? [2]
IQR = Q₃ − Q₁ is the spread of the middle half of the data. It ignores the extreme values, so one unusual value does not change it much.
- How do you find the median from a cumulative frequency curve? [2]
Go across from n ÷ 2 on the cumulative frequency axis to the curve, then down to the horizontal axis and read the value. Use n ÷ 4 and 3n ÷ 4 for the quartiles.
- What does r = 0.9 mean for height and arm span? [2]
There is a strong positive correlation: taller learners usually have longer arm spans, and the points lie close to the line of best fit.
- The marks of 10 learners are 12, 15, 15, 18, 20, 21, 24, 25, 30, 45. (a) Find the mean, median and mode. (b) Find the quartiles and the interquartile range. (c) Which average best represents the marks? Give a reason. [7]
(a) Mean = 225 ÷ 10 = 22.5 (1); median = (20 + 21) ÷ 2 = 20.5 (1); mode = 15 (1). (b) Q₁ = 15, Q₃ = 25 (1, 1); IQR = 10 (1). (c) The median, because the outlier 45 pulls the mean up (1).
- Calculate the standard deviation of 2, 4, 4, 4, 5, 5, 7, 9. [3]
Mean = 40 ÷ 8 = 5 (1); Σ(x − x̄)² = 9 + 1 + 1 + 1 + 0 + 0 + 4 + 16 = 32 (1); σ = √(32 ÷ 8) = √4 = 2 (1).
- Heights (cm) of 40 learners: 140–145: 3; 145–150: 9; 150–155: 14; 155–160: 10; 160–165: 4. (a) Draw a cumulative frequency table. (b) Estimate the median from a cumulative frequency curve. (c) Estimate the mean using mid-points. [7]
(a) Upper boundaries 145, 150, 155, 160, 165 with cumulative frequencies 3, 12, 26, 36, 40 (2). (b) Median at the 20th value: 150 + (20 − 12) ÷ 14 × 5 ≈ 152.9 cm (2). (c) Σfx = 3 × 142.5 + 9 × 147.5 + 14 × 152.5 + 10 × 157.5 + 4 × 162.5 = 6 120 (1); mean = 6 120 ÷ 40 = 153 cm (2).
- A ruler is caught after falling 18 cm. Calculate the reaction time. (g = 9.8 m/s²) [2]
d = ½gt², so t = √(2d ÷ g) = √(2 × 0.18 ÷ 9.8) (1) = 0.19 s (1).