Year 10 Extended · Unit 6 Statistics and the bell curve · Lesson 4 of 7

Two routes to the same spread

A frequency table repeats each value, so each squared deviation is counted f times. This lesson finds the standard deviation from tables, and tests a second form of the formula against the first.

Statistics and ProbabilityGDC optional

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Lesson 3: a set of 4 values has Σ(x − μ)² = 36. What is the standard deviation?
    Answer3. The variance is 36 ÷ 4 = 9 and √9 = 3.
  2. From Year 9: expand (x − 5)².
    Answerx² − 10x + 25. The middle term is −2 × 5 × x.
  3. From Lesson 1: a table has 4 values of 2, 6 values of 5 and 2 values of 8. What is the mean?
    Answer4.5. The total is 8 + 30 + 16 = 54 and 54 ÷ 12 = 4.5.
Words used in this lesson
frequency form
the standard deviation with each squared deviation counted f times. For example: σ² = Σf(x − μ)² ÷ n.
summary form
a second way to write the variance. For example: it uses Σfx and Σfx², not deviations.
mean of the squares
Σfx² divided by n. For example: (1 + 9) ÷ 2 = 5.
square of the mean
the mean multiplied by itself. For example: the mean 1.5 gives 2.25.
grouped table
a table of classes with a frequency each. For example: 0 ≤ x < 10: 4 values.
total frequency
the sum of all the frequencies, n. For example: 3 + 5 + 2 = 10.

The library tally

A library in Seoul tallies how many books each visitor borrows in one afternoon: 20 visitors, borrowing 0 to 4 books. Instead of 20 separate values, there is a frequency table. Repeating each deviation f times is slow, and the mean is not a whole number, so every deviation is a decimal.

Predict first: is there a way to find the spread that never subtracts the mean from a value? Try both forms in the figure and see whether they ever disagree.

1
1
3mean
1.3variance
1.14standard deviation

Table 1, deviation form: Σf(x − μ)² ÷ n = 26 ÷ 20 = 1.3, so the standard deviation is 1.14.

Show me the steps
  1. Keep table 1 with the deviation form. The deviations are −2, −1, 0, 1, 2 with frequencies 2, 5, 6, 5, 2, so Σf(x − μ)² = 8 + 5 + 0 + 5 + 8 = 26. Then 26 ÷ 20 = 1.3 and σ = 1.14.
  2. Switch to the summary form on table 1. Σfx² = 206, and 206 ÷ 20 = 10.3. The mean is 3, so subtract 3² = 9: 10.3 − 9 = 1.3. The same variance, with no deviations at all.
  3. Choose table 2 with the summary form. The mean is 1.4, which would make 20 awkward deviations. Σfx² = 66, so 66 ÷ 20 = 3.3, and 3.3 − 1.4² = 3.3 − 1.96 = 1.34. The standard deviation is 1.16.
  4. Choose table 3 with the deviation form. Σf(x − μ)² = 4 + 4 + 0 + 4 + 4 = 16, and 16 ÷ 20 = 0.8, giving σ = 0.89. Switch the form and 16.8 − 16 = 0.8 again.

Frequency form. With n = Σf, the mean is μ = Σfx ÷ n and σ = √(Σf(x − μ)² ÷ n). Each squared deviation is counted f times.

Summary form. σ = √(Σfx² ÷ n − μ²): the mean of the squares minus the square of the mean. It needs only Σfx, Σfx² and n, so no deviation is ever worked out.

Grouped data. Use the mid-interval values as x. The standard deviation found is an estimate, because the real values inside a class are not known.

fx² means f × x², not (fx)². Square the value, then multiply by the frequency. For x = 3 with f = 4, fx² = 4 × 9 = 36. The wrong reading (4 × 3)² = 144 is four times too big.

Worked examples

1. Books borrowed: 0 books by 4 visitors, 1 by 8, 2 by 6, 3 by 2. Find σ using deviations.

  1. n = 20 and Σfx = 0 + 8 + 12 + 6 = 26, so μ = 1.3.
  2. Σf(x − μ)² = 4(1.69) + 8(0.09) + 6(0.49) + 2(2.89) = 6.76 + 0.72 + 2.94 + 5.78 = 16.2.
  3. The variance is 16.2 ÷ 20 = 0.81, and σ = √0.81 = 0.9.

2. The same table with the summary form.

  1. Σfx² = 4(0) + 8(1) + 6(4) + 2(9) = 50, so Σfx² ÷ n = 2.5.
  2. μ² = 1.3² = 1.69, so the variance is 2.5 − 1.69 = 0.81: the same, with no deviations.

3. Why the two forms always agree.

  1. Expand: Σ(x − μ)² = Σx² − 2μΣx + nμ². Since Σx = nμ, this is Σx² − 2nμ² + nμ² = Σx² − nμ².
  2. Divide by n: the variance is Σx² ÷ n − μ². The forms are the same quantity, written two ways.

Your go. The first step is given. A grouped table has 0 to under 20: 3 values; 20 to under 40: 9; 40 to under 60: 6; 60 to under 80: 2. Estimate the standard deviation to one decimal place, using the summary form.

  1. The mid-interval values are 10, 30, 50 and 70, and n = 20. Then Σfx = 30 + 270 + 300 + 140 = 740, so μ = 37.
  2. Now find Σfx², divide by n, subtract μ², and take the root.
Check the answer

Σfx² = 3(100) + 9(900) + 6(2500) + 2(4900) = 33 200, and 33 200 ÷ 20 = 1660. Then 1660 − 37² = 1660 − 1369 = 291, and σ = √291 = 17.1.

With a GDC (optional)

Put the values in one list and the frequencies in another, and ask for the one-variable statistics with the frequency list named. Read the standard deviation that divides by n. For a grouped table, use the mid-interval values as the data; the result is then an estimate of the spread.

The GDC reports the quantities Σx and Σx² as well, which are exactly what the summary form needs. Everything here can be done by hand.

Check before you go on

1. A data set has n = 10, Σx² = 340 and mean 5. Which calculation gives its standard deviation by the summary form?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. A table has x = 1, 2, 3 with frequencies 2, 5, 3. The mean is 2.1. Find the standard deviation, to one decimal place.

3. A table has n = 20, Σfx = 80 and Σfx² = 400. Find the standard deviation.

Level 2 · Building

4. A grouped table has 0 to under 10: 4 values; 10 to under 20: 10; 20 to under 30: 6. Using mid-interval values, find the estimated standard deviation.

5. A table has n = 20 and mean 7, and its standard deviation is 3. What is Σfx²?

Level 3 · Stretch

6. For 6 values, Σx = 42 and Σ(x − μ)² = 30. Find Σx².

7. The values x = 1, 3 and 5 have frequencies 4, f and 4. The mean is 3, and the variance is 3.2. Find f.

Something to think about

The summary form says variance = (mean of the squares) − (square of the mean). Use it to explain why the variance can never be negative, and what must be true of the data when the variance is exactly 0.

One way to think about it

The variance is a mean of squares of deviations, so it cannot be negative; equivalently, the mean of the squares can never be less than the square of the mean. It is exactly 0 only when every deviation is 0, which means every value equals the mean. Then the mean of the squares and the square of the mean are equal.

Exit ticket

8. Every frequency in a table is doubled, and the x values stay the same. What happens to the standard deviation?