3 mistakes learners make with averages from a frequency table, each one named and explained.
Each value counts once for every time it occurred, which is what the frequency column is for.
The total is worked out correctly and then divided by how many rows the table has.
The rows are what you can see and count, and the real count is hidden in a column.
Question. Four rows, twenty pupils
A common answer. 128 / 4 = 32
The answer. 128 / 20 = 6.4
Why. You are averaging over pupils, and there are twenty of those.
Add the frequency column up first. That number is what you divide by.
The distinct values are averaged as if each occurred once, so the frequencies do nothing.
The values are in a neat column and look exactly like a small data set on their own.
Question. Sizes 5, 6, 7, 8 with frequencies 4, 7, 6, 3
A common answer. (5 + 6 + 7 + 8) / 4 = 6.5
The answer. 128 / 20 = 6.4
Why. Seven pupils wear size 6, so size 6 has to count seven times.
The frequency says how many times to count each value. Every pupil counts once.
Grouped data is estimated using the highest value in each class, so the estimate comes out high.
The upper bound is the number printed at the end of the interval, so it is the one you see.
Question. 0 < t <= 10 with 8 people
A common answer. using 10 as the value
The answer. using 5, the midpoint
Why. Those eight people are spread across the whole interval, so the middle is the fair estimate.
It is an estimate because you no longer know the values. The midpoint is the fairest guess.
WAJD spots these patterns in your child's answers and names the one behind their wrong answers, instead of just marking them wrong.