B. Probability Distributions · Continuous Distributions and the Normal Distribution

Mean and Standard Deviation of Grouped Continuous Data

B2.2 B2.4About 65 minutes

Learning goals

I can find the midpoint of each class interval in a frequency table.
I can estimate the mean of grouped continuous data using Σfm ÷ Σf.
I can estimate the standard deviation of grouped data, for a population or a sample.
I can explain why grouped results are estimates and how class width affects them.

Words to know

grouped data
Data organized into class intervals with a frequency for each interval, instead of a list of individual values.

class midpoint (m)
The middle of a class interval: (lower boundary + upper boundary) ÷ 2. It stands in for every value in that class.

frequency (f)
The number of data values in a class.

grouped mean
An estimate of the mean from a frequency table: x̄ ≈ Σfm ÷ Σf.

grouped standard deviation
An estimate of the standard deviation from a frequency table: σ ≈ √(Σf(m − μ)² ÷ N), or s ≈ √(Σf(m − x̄)² ÷ (n − 1)) for a sample.

When you only have the table

Statistics Canada, sports leagues and school boards often publish continuous data already grouped into a frequency table. You no longer have the individual values. Can you still find the mean and the standard deviation? Yes, approximately.

The trick: inside each class, pretend that every value sits exactly at the class midpoint m. Values below the midpoint and above it roughly balance, so the estimate is usually close.

Nine run times between 5.0 and 6.0 min are all treated as if they were 5.5 min, the midpoint of the class. Some were lower and some were higher, so the errors partly cancel.

The grouped formulas

For each class, multiply the midpoint by the frequency. Then:

Mean: μ (or x̄) ≈ Σfm ÷ Σf
Standard deviation (population): σ ≈ √( Σf(m − μ)² ÷ N )
Standard deviation (sample): s ≈ √( Σf(m − x̄)² ÷ (n − 1) )

Here N (or n) = Σf, the total frequency. The shortcut form from the last lesson also works and avoids messy decimals: Σf(m − x̄)² = Σfm² − n·x̄².

Each midpoint is counted f times, which is why every sum has an f in it.

Worked example

Example 1: Grouped mean and SD of run times

The 1 km run times of a class of 40 students (from the first lesson of this unit) are below. Treat the class as the population. Estimate μ and σ.

Midpoints: 4.5, 5.5, 6.5, 7.5, 8.5. Frequencies: 3, 9, 14, 10, 4. N = 40.
fm: 13.5, 49.5, 91, 75, 34. Σfm = 263. So μ ≈ 263 ÷ 40 = 6.575 min.
fm²: 3(20.25) = 60.75, 9(30.25) = 272.25, 14(42.25) = 591.5, 10(56.25) = 562.5, 4(72.25) = 289. Σfm² = 1776.
Σf(m − μ)² = Σfm² − Nμ² = 1776 − 40(6.575)² = 1776 − 1729.225 = 46.775.
σ² ≈ 46.775 ÷ 40 = 1.1694, so σ ≈ √1.1694 ≈ 1.08.

μ ≈ 6.58 min and σ ≈ 1.08 min.

Time (min)	m	f	fm	fm²
4 to 5	4.5	3	13.5	60.75
5 to 6	5.5	9	49.5	272.25
6 to 7	6.5	14	91	591.5
7 to 8	7.5	10	75	562.5
8 to 9	8.5	4	34	289
Total		40	263	1776

Work table for Example 1. Add columns fm and fm² and total them.

Worked example

Example 2: A sample of pumpkins

A pick-your-own farm weighs a random sample of 50 pumpkins in October. Estimate the sample mean and sample standard deviation.

2 to 4 kg: 6 pumpkins
4 to 6 kg: 12 pumpkins
6 to 8 kg: 17 pumpkins
8 to 10 kg: 10 pumpkins
10 to 12 kg: 5 pumpkins

Midpoints: 3, 5, 7, 9, 11. n = 6 + 12 + 17 + 10 + 5 = 50.
Σfm = 18 + 60 + 119 + 90 + 55 = 342. x̄ ≈ 342 ÷ 50 = 6.84 kg.
Σfm² = 6(9) + 12(25) + 17(49) + 10(81) + 5(121) = 54 + 300 + 833 + 810 + 605 = 2602.
Σf(m − x̄)² = 2602 − 50(6.84)² = 2602 − 2339.28 = 262.72.
It is a sample, so divide by n − 1 = 49: s² ≈ 262.72 ÷ 49 ≈ 5.3616, and s ≈ 2.32.

x̄ ≈ 6.84 kg and s ≈ 2.32 kg.

The pumpkin sample. Dots mark the midpoints used in the calculation. The dashed line is the mean, 6.84 kg, and the bracket shows one standard deviation (about 2.32 kg) on each side of it.

Why the answers are only estimates

The grouped mean would be exact only if the values in each class were centred on the midpoint. With narrower classes the midpoints are closer to the real values, so the estimates improve; with very wide classes they can be off. If you have the raw data, use it. In the next lesson you will see how class width also changes the shape of a histogram.

Worked example

Example 3: Grouped versus raw

Six test scores are 41, 44, 47, 52, 55 and 58. (a) Find the exact mean. (b) Group them into the classes 40 to 50 and 50 to 60 and estimate the mean. (c) Explain the difference.

(a) Exact: (41 + 44 + 47 + 52 + 55 + 58) ÷ 6 = 297 ÷ 6 = 49.5.
(b) Three scores in each class. Grouped: (45 × 3 + 55 × 3) ÷ 6 = 300 ÷ 6 = 50.
(c) In the 40 to 50 class the real scores (41, 44, 47) average 44, a little below the midpoint 45, and in the 50 to 60 class they average 55, exactly at the midpoint. So the grouped mean comes out 0.5 too high.

Exact 49.5; grouped estimate 50. The small error comes from treating each score as its class midpoint.

Just for fun

Grouped data's team motto: "We're all in this interval together." The midpoint is the team captain who speaks for everyone.

Hands-on activity

Group your own data

You need: your 20 alphabet times from the first lesson (or 20 new times), a calculator.

Choose a class width (0.2 s or 0.5 s works well) and make a frequency table with columns class, m, f, fm and fm².
Estimate x̄ and s from the table.
Now find x̄ and s from the 20 raw times with your calculator's STAT mode.
Repeat with a different class width. Which width gave estimates closer to the raw values?

What to look for: the grouped estimates should be close to the raw values (usually within a few hundredths of a second for the mean). Narrower classes usually give closer estimates.

Your turn

What is the midpoint of the class 2.5 to 3.0 s?

Shovelling times (min) for 20 households after a storm are grouped: 10 to 20 (4), 20 to 30 (10), 30 to 40 (5), 40 to 50 (1). Treat them as a population. Estimate μ.

For the same shovelling data, estimate σ.

A sample of 20 icicle lengths (cm): 10 to 15 (2), 15 to 20 (7), 20 to 25 (8), 25 to 30 (3). Estimate x̄ and s.

In the run-time table (40 students), which class contains the median?

Why is the grouped mean only an estimate?

More practice: warm-up

What is the midpoint of the class 40 to 50?

In a frequency table the frequencies are 3, 7, 12, 6 and 2. How many data values are there?

A class has midpoint 12.5 and frequency 6. What is fm?

In a grouped table, which value stands in for every data value in a class?

Σfm = 480 and Σf = 40. Estimate the mean.

More practice: on your own

A sample of 20 students' daily screen time (h): 0 to 2 (5), 2 to 4 (8), 4 to 6 (5), 6 to 8 (2). Estimate x̄ and s.

Classes with midpoints 5, 15 and 25 have frequencies 4, k and 6. The grouped mean is 16. Find k.

The pumpkin sample had s ≈ 2.32 kg. In one sentence, explain what this tells a farmer.

Two grouped tables summarize the same 200 values: one uses class width 1, the other class width 10. Which is likely to give a more accurate estimate of the mean? Why?

For the run-time data, μ ≈ 6.575 min. Use the table to find Σf(m − μ) and explain the result.

Challenge

For the run-time class (μ ≈ 6.575, σ ≈ 1.08), estimate how many of the 40 students ran within one standard deviation of the mean. Assume times are spread evenly within each class.

Two more students join the run-time class, both in the class 9 to 10 min. Without full calculation, predict whether μ and σ increase or decrease. Then check.

Explain why Σf(m − x̄)² = Σfm² − n·x̄² for grouped data, where n = Σf.

Check your understanding

Quiz

What is the midpoint of the class 15 to 20?

A table has Σf = 25 and Σfm = 350. What is the estimated mean?

Why is a mean found from grouped data called an estimate?

For a sample of grouped data, what do you divide Σf(m − x̄)² by before taking the square root?

Σfm² = 2602, n = 50 and x̄ = 6.84. What is Σf(m − x̄)²?

Which change usually makes grouped estimates of the mean and SD more accurate?

For parents and teachers

Big idea (B2.2, B2.4): when continuous data come in a frequency table, the mean and standard deviation can be estimated by treating every value in a class as the class midpoint and weighting by frequency.

How to help: Ask your teen to set up the work table (m, f, fm, fm²) for one example and explain each column. A spreadsheet is a great check: type the midpoints and frequencies, then compute =SUMPRODUCT(B2:B6,C2:C6)/SUM(C2:C6) for the mean.

Common difficulties:

Dividing Σfm by the number of classes instead of by Σf (the number of values).
Forgetting the f when squaring: it is f × m², not (f × m)².
Using class boundaries instead of midpoints.
Using n instead of n − 1 for a sample (or the reverse).