Standard Deviation: Grouped

A frequency table: each value counts f times

A team scores 1 goal in 3 games, 2 goals in 5 games and 3 goals in 2 games. Written out, the ten values are 1, 1, 1, 2, 2, 2, 2, 2, 3, 3. With x for a value and f for its frequency, the mean is $\sum fx÷\sum f=\left(1\times 3+2\times 5+3\times 2\right)/10=19/10=1.9$Σ f x ÷ Σ f = (1 × 3 + 2 × 5 + 3 × 2) / 10 = 19 / 10 = 1.9 goals.

The standard deviation of the ten values needs the square of every value, and the value 1 occurs 3 times, so its square counts 3 times: $1^2\times 3=3$1² × 3 = 3. In the same way, $2^2\times 5=20$2² × 5 = 20 and $3^2\times 2=18$3² × 2 = 18. So $\sum f{x}^2=3+20+18=41$Σ f x² = 3 + 20 + 18 = 41, the sum of the squares of all ten values.

The frequencies act as weights, and the shorter formula becomes $\sigma =\sqrt{\frac{\sum f{x}^2}{\sum f}-{\bar{x}}^2}$σ = √(Σ f x²/Σ f − x̄²). Here the variance is $41/10-1.9^2=4.1-3.61=0.49$41 / 10 − 1.9² = 4.1 − 3.61 = 0.49, and $\sigma =\sqrt{0.49}=0.7$σ = √0.49 = 0.7 goals.

The totals give the mean, 19 ÷ 10 = 1.9, and the variance, $41/10-1.9^2=0.49$41 / 10 − 1.9² = 0.49. The standard deviation is $\sqrt{0.49}=0.7$√0.49 = 0.7 goals.

Grouped data keeps only the counts

Ten plants are measured, and their heights are recorded only by class. Writing h for a height in centimeters, 2 plants are in the class $0\le h<10$0 ≤ h < 10, 6 are in $10\le h<20$10 ≤ h < 20 and 2 are in $20\le h<30$20 ≤ h < 30. The middle class holds most of the plants.

The heights themselves are gone, so there are no values to square. As with the mean of grouped data, each class needs one value to stand in for all of its values.

The ten heights by class: 2, then 6, then 2 plants.

Midpoints, weighted by the counts

Use the midpoint of each class, halfway between its two ends: 5, 15 and 25 cm. Treat the 2 plants in the first class as 5 cm tall, the 6 in the second as 15 cm, and the 2 in the third as 25 cm, and weight each midpoint by its frequency.

The estimated mean is $\sum fx÷\sum f=\left(5\times 2+15\times 6+25\times 2\right)/10=\left(10+90+50\right)/10=150/10=15$Σ f x ÷ Σ f = (5 × 2 + 15 × 6 + 25 × 2) / 10 = (10 + 90 + 50) / 10 = 150 / 10 = 15 cm. From here the method is the same as for a frequency table, with the midpoints as the values.

The midpoints 5, 15 and 25 sit halfway along their classes. The estimated mean, 15, is the middle midpoint, since the first and last classes hold 2 plants each.

Each class adds its squared distance, f times

Take the classes one at a time. The 2 plants at midpoint 5 are 15 − 5 = 10 from the mean. Each adds $10^2=100$10² = 100 to the total of the squared differences, so together they add $2\times 10^2=200$2 × 10² = 200.

The 6 plants at midpoint 15 are at the mean, so they add $6\times 0^2=0$6 × 0² = 0. The 2 plants at midpoint 25 are 10 above the mean and add $2\times 10^2=200$2 × 10² = 200. Altogether, $\sum f{\left(x-\bar{x}\right)}^2=200+0+200=400$Σ f(x − x̄)² = 200 + 0 + 200 = 400.

Share that total over the 10 plants: 400 ÷ 10 = 40 is the variance. Its square root is $\sqrt{40}\approx 6.3$√40 ≈ 6.3, so the standard deviation of the heights is about 6.3 cm.

The shorter formula gives the same answer: $\sum f{x}^2=5^2\times 2+15^2\times 6+25^2\times 2=50+1350+1250=2650$Σ f x² = 5² × 2 + 15² × 6 + 25² × 2 = 50 + 1350 + 1250 = 2650, and $2650/10-15^2=265-225=40$2650 / 10 − 15² = 265 − 225 = 40.

A full table

Twenty students record the hours h they spend on homework in one week: 2 are in the class $0\le h<2$0 ≤ h < 2, 6 in $2\le h<4$2 ≤ h < 4, 8 in $4\le h<6$4 ≤ h < 6 and 4 in $6\le h<8$6 ≤ h < 8. Estimate the mean and the standard deviation.

Set out a table with a row for each class and four columns: the midpoint x, the frequency f, the product fx and the product fx². The midpoints are 1, 3, 5 and 7. The fx column is 2, 18, 40 and 28, and the fx² column is $1^2\times 2=2$1² × 2 = 2, $3^2\times 6=54$3² × 6 = 54, $5^2\times 8=200$5² × 8 = 200 and $7^2\times 4=196$7² × 4 = 196. The totals are $\sum f=20$Σ f = 20, $\sum fx=88$Σ f x = 88 and $\sum f{x}^2=452$Σ f x² = 452.

The estimated mean is $\sum fx÷\sum f=88/20=4.4$Σ f x ÷ Σ f = 88 / 20 = 4.4 hours. The variance is $\sum f{x}^2÷\sum f-{\bar{x}}^2=452/20-4.4^2=22.6-19.36=3.24$Σ f x² ÷ Σ f − x̄² = 452 / 20 − 4.4² = 22.6 − 19.36 = 3.24, and the standard deviation is $\sqrt{3.24}=1.8$√3.24 = 1.8 hours.

The definition gives the same answer with more work. The midpoints are −3.4, −1.4, +0.6 and +2.6 from the mean, their squares are 11.56, 1.96, 0.36 and 6.76, and weighted by the frequencies these give 23.12, 11.76, 2.88 and 27.04, a total of 64.8. Then 64.8 ÷ 20 = 3.24, as before.

The totals give the estimated mean, 88 ÷ 20 = 4.4 hours, and the variance, $452/20-4.4^2=3.24$452 / 20 − 4.4² = 3.24. The standard deviation is $\sqrt{3.24}=1.8$√3.24 = 1.8 hours.

An estimate, like the grouped mean

A standard deviation from grouped data is an estimate. Every value in a class was taken to be at its midpoint, but the real values are spread through the class. Say that the standard deviation of the homework hours is about 1.8 hours, and that it is an estimate.

Two slips are common. The first is to leave out the frequencies and average the squared distances of the midpoints alone, which counts a class of 2 students as much as a class of 8. The second is to use the upper end of each class, or the class width, in place of the midpoint.