A class such as 10 ≤ x < 20 tells us an observation belongs between ten inclusive and twenty exclusive. It does not reveal whether that observation was 11, 15 or 19.8. Frequency five says there are five observations in that interval, not five copies of the upper boundary.

Once exact values are hidden, the original sum is generally unknown. Many different data sets can fit the same grouped table. The midpoint calculation creates a useful representative total; it does not magically reconstruct the lost measurements. This explains why an estimated mean should retain the word estimated.

Average the class boundaries: midpoint = (lower + upper)/2. For 0 ≤ x < 10 the midpoint is five; for 10 ≤ x < 20 it is fifteen; and for 20 ≤ x < 30 it is twenty-five. These are the values used in the standard midpoint approximation.

An open upper endpoint does not change the midpoint formula. But interpret the class as supplied. A class of integer scores 10–19 has midpoint 14.5, not fifteen. Do not silently rewrite the intervals into different boundaries. Read whether the table concerns continuous measurements, integer values or explicitly stated class boundaries.

Represent the two observations in the first class by two fives, giving ten. Represent the five observations in the second class by five fifteens, giving seventy-five. Represent the three observations in the third class by three twenty-fives, giving seventy-five. The estimated total is 160.

The count is 2 + 5 + 3 = 10. Therefore estimated mean = 160/10 = 16. The denominator counts observations, not the three class rows. Each frequency has two jobs in the table: it weights a representative value in the numerator and contributes to the observation count in the denominator.

Using ten, twenty and thirty gives total 2 × 10 + 5 × 20 + 3 × 30 = 210, divided by ten to give 21. For these lower-inclusive, upper-exclusive intervals, every actual observation is below its class’s upper boundary. Hence the true mean is below 21.

That limiting value is not the standard midpoint estimate. Using lower boundaries instead gives total 110 and mean eleven, which is a lower bound. Thus the grouped data imply 11 ≤ true mean < 21 under the stated continuous interval model. The midpoint estimate sixteen sits between them but is not guaranteed to equal the true mean.

Yes. The values 1, 2, 11, 12, 13, 14, 15, 21, 22 and 23 fit the three frequencies and have sum 134, giving mean 13.4. The values 8, 9, 15, 16, 17, 18, 19, 27, 28 and 29 also fit and have sum 186, giving mean 18.6.

Both tables of exact values collapse to the same grouped counts, so both produce midpoint estimate sixteen. The example demonstrates lost information, not an arithmetic failure. Avoid claiming that the observations are known to be uniformly distributed within each class. Midpoint representation is an approximation whose error depends on the hidden values.

No. If the frequencies are supplied, use each class’s own midpoint and multiply by its frequency. For 0 ≤ x < 10 with frequency 2 and 10 ≤ x < 30 with frequency 3, midpoints are five and twenty. The estimate is (10 + 60)/5 = 14.

Do not multiply by class width again when the column already gives frequencies. In a histogram, a height may instead represent frequency density, so frequency must first be recovered from density × class width. Identify what the supplied column or axis measures before building the mean calculation. Different graph quantities are not interchangeable.

Question 1: classes 0 ≤ x < 4 and 4 ≤ x < 8 have frequencies 3 and 1; estimate the mean. Question 2: classes 10 ≤ x < 20 and 20 ≤ x < 40 have frequencies 4 and 2; estimate the mean. Question 3: five observations lie in 6 ≤ x < 10; is their mean known exactly?

Question 4: an exact-value table gives x = 2 with frequency 3 and x = 6 with frequency 1; find the mean. Answers: 1) (3 × 2 + 1 × 6)/4 = 3. 2) (4 × 15 + 2 × 30)/6 = 20. 3) No; the midpoint estimate is eight. 4) Three, exactly from the stated exact values.

Ask the student to label the class boundaries, midpoint, frequency and product separately. A four-column table can reveal a wrong representative value before the final division. Bring the original grouped table and working to tuition; ‘uses upper boundary as every value’ is a more precise learning note than ‘weak in averages’.

In a small-group 3-pax lesson, students can build different raw data sets satisfying the same grouped counts and compare their true means with one shared midpoint estimate. For parents considering Secondary 4 Mathematics tuition in Punggol, the useful checkpoint is whether your child can say which information is known and which has been approximated. That distinction makes the final answer honest and easier to check.