Egypt Preparatory 2 Mathematics, Week 14: Mean of a grouped table: class midpoints and f × x

Mean of a grouped table: class midpoints and f × x

Mathematics Prep 2 Term 1, Unit 4, Lesson 2

GoalFind the mean of a grouped table with the class midpoints, the products f × m and the totals, and find an unknown frequency from a given mean.

New words

midpoint m · (lower limit + upper limit) ÷ 2; it stands for every value in the classf × m · the frequency times the midpoint: an estimate of the total of the classmean · total of f × m divided by the total of festimate · grouped data hide the real values, so the mean is not exact

Explanation

In a grouped table we do not know the real values, so we let the midpoint of each class stand for all the values in it. The midpoint is (lower limit + upper limit) ÷ 2: for the class 10– (10 up to 20) it is (10 + 20) ÷ 2 = 15. The heights of 20 seedlings are in the table. Add a column m for the midpoints and a column f × m. Then mean = (total of f × m) ÷ (total of f). Here the total of f × m is 30 + 125 + 280 + 225 = 660 and the total of f is 20, so the mean is 660 ÷ 20 = 33 cm. It is an estimate. Divide by the total of the frequencies (20), not by the number of classes (4). If a frequency k is unknown, write the total of f × m and the total of f with k, put them in the mean formula and solve the equation.

Class	f	midpoint m	f × m
10–	2	15	30
20–	5	25	125
30–	8	35	280
40–	5	45	225
Total	20		660

Worked examples

Mean of the heights of 20 seedlings (the table above): the midpoints are 15, 25, 35, 45; f × m gives 30, 125, 280, 225, total 660; total of f is 20. Mean = 660 ÷ 20 = 33 cm.

Class	f	midpoint m	f × m
60–	3	65	195
70–	9	75	675
80–	12	85	1020
90–	6	95	570
Total	30		2460

Speeds of 30 cars in the classes 60–, 70–, 80–, 90–100 km/h with frequencies 3, 9, 12, 6. The midpoints are 65, 75, 85, 95. f × m: 195, 675, 1020, 570, total 2460. Mean = 2460 ÷ 30 = 82 km/h.

An unknown frequency. Classes 0–, 10–, 20–, 30–40 have frequencies 4, k, 10, 6 and the mean is 21. Midpoints 5, 15, 25, 35. Total of f × m = 20 + 15k + 250 + 210 = 480 + 15k and total of f = 20 + k. So (480 + 15k) ÷ (20 + k) = 21, 480 + 15k = 420 + 21k, 60 = 6k and k = 10. Check: total of f = 30, total of f × m = 630, 630 ÷ 30 = 21.

Class activity

Class mean. Measure the height of 20 classmates in cm, group them in classes of 10 cm and find the mean with the midpoints. Then find the exact mean from the raw heights and compare. How close is the grouped estimate?

Practice

Classes 0–, 10–, 20–, 30–40 with frequencies 3, 7, 6, 4. Find the mean.

Find the midpoints of the classes 80– (to 90) and 20– (to 30).

Classes 0–, 5–, 10–, 15–20 have frequencies 2, k, 6, 2 and the mean is 9.5. Find k. (Midpoints 2.5, 7.5, 12.5, 17.5.)

Marks in the classes 0–, 10–, 20–, 30–40 have frequencies 5, 10, 15, 10. Find the mean mark.

Sami finds the total of f × m is 660 and divides by the number of classes, 4, to get 165. What went wrong?

Mean, median and mode

Mean of a grouped table: class midpoints and f × x