The total frequency of this table is 60. 0 < t 10: 9 10 < t 20: x 20 < t 30: 21 30 < t 40: 12 Work out the value of x, and write down the midpoint of the class that contains the median.

x = 18, midpoint of the median class = 25

Two frequency polygons are drawn on the same axes for two groups of 60 people, using the same classes. Polygon P has its highest point at (15, 22). Polygon Q has its highest point at (35, 22). Which statement is correct?

Q has a higher modal class than P, so the values in group Q are typically larger

The two groups have the same modal class

Group P contains more people than group Q

Nothing can be compared, because a frequency polygon does not show frequencies

Why is the mean from a grouped frequency table only an estimate?

Because the table records which class each value fell into and nothing more, so the exact total of the data cannot be recovered. Using the midpoint of a class assumes every value in it sits at the centre, which is almost never true. That assumption is what makes the answer an estimate, and it is why Edexcel writes the instruction as work out an estimate for the mean rather than calculate the mean.

Are grouped frequency tables on the Foundation paper?

Yes. The estimated mean, the modal class and the class containing the median are all crossover content on Edexcel 1MA1, so a Foundation and a Higher paper in the same series can carry very similar questions. The Higher version tends to add a second demand on top, such as a missing frequency to find or a comparison of two distributions to write.

Where do the points go on a frequency polygon?

At the midpoint of each class, at a height equal to the frequency of that class, joined in order with straight line segments. A class of 20 < t 30 with frequency 14 is plotted at 25 across and 14 up. Points plotted at the ends of the intervals, and outlines closed back to the first point, are the two errors examiners report on this part.

How many marks is an estimated mean question worth on Edexcel Higher?

Three marks in the standard form: one for the fx products, one dependent mark for dividing by the total frequency, and one for the answer. With a modal class part and a frequency polygon or a comparison alongside it, the whole question usually comes to 3-5 marks.

Grouped Frequency Tables - Edexcel GCSE Higher Maths

Q: Why is the mean from a grouped frequency table only an estimate?

Because the table records which class each value fell into and nothing more, so the exact total of the data cannot be recovered. Using the midpoint of a class assumes every value in it sits at the centre, which is almost never true. That assumption is what makes the answer an estimate, and it is why Edexcel writes the instruction as work out an estimate for the mean rather than calculate the mean.

Q: Are grouped frequency tables on the Foundation paper?

Yes. The estimated mean, the modal class and the class containing the median are all crossover content on Edexcel 1MA1, so a Foundation and a Higher paper in the same series can carry very similar questions. The Higher version tends to add a second demand on top, such as a missing frequency to find or a comparison of two distributions to write.

Q: Where do the points go on a frequency polygon?

At the midpoint of each class, at a height equal to the frequency of that class, joined in order with straight line segments. A class of 20 < t 30 with frequency 14 is plotted at 25 across and 14 up. Points plotted at the ends of the intervals, and outlines closed back to the first point, are the two errors examiners report on this part.

Q: How many marks is an estimated mean question worth on Edexcel Higher?

Three marks in the standard form: one for the fx products, one dependent mark for dividing by the total frequency, and one for the answer. With a modal class part and a frequency polygon or a comparison alongside it, the whole question usually comes to 3-5 marks.

Typically 3-5 marks, usually in one structured question · Spec 1MA1

Grouping data throws the individual values away. A table saying that 14 people took between 20 and 30 minutes does not say whether those 14 took 21 minutes or 29, so every average has to be downgraded: the mode becomes a modal class, the median becomes the class the median falls in, and the mean becomes an estimate built from the midpoint of each class. On Edexcel 1MA1 this sits under spec ref S4, and it is crossover content, so the same table can be set at the end of a Foundation paper and near the start of a Higher one.

Edexcel sets it as one question in two or three short parts. Writing down the modal class is worth a mark. Working out an estimate for the mean is worth three, and the mark scheme has a shape worth learning: the first method mark is for products $fx$ using a value taken consistently from inside each interval, the second is dependent on the first and is for dividing the total of those products by the total frequency, and the accuracy mark needs the right figure. A candidate who uses the top end of every class instead of the midpoint still banks two of the three marks, which is why the $fx$ column earns more than the answer line does. The question turns up on all three papers, and on Paper 1 the frequencies are chosen so the multiplications can be done by hand.

Three slips account for most of the lost marks. The first is the divisor: dividing the total of the $fx$ column by the number of classes rather than by the total frequency turns 21.5 minutes into 215, and an average sitting outside every class in the table should stop you before an examiner does. The second is answering the modal class with a frequency, writing 19 when the answer is an interval, or picking the middle row because it looks central. The third is the frequency polygon, which goes at the midpoints, is joined with straight line segments, and is left open rather than closed back round to the first point.

Worked example

The table gives information about the times, in minutes, taken by 40 people to finish a crossword. Time

0 < t \le 10

: 6 people Time

10 < t \le 20

: 11 people Time

20 < t \le 30

: 14 people Time

30 < t \le 40

: 9 people (a) Write down the modal class. (b) Work out an estimate for the mean time. (c) Write down the class that contains the median.

[5 marks]

For (a), the largest frequency in the table is 14, and it belongs to the class $20 < t \le 30$ . That interval is the modal class.
B1: the interval itself; an answer of 14 is the frequency, not the class, and scores nothing
For (b), write a midpoint beside each row. Add the two boundaries and halve: $\frac{0 + 10}{2} = 5$ , then 15, then 25, then 35.
M1: midpoints, or any value taken consistently from inside each interval
Multiply each midpoint by its frequency and total the column: $5 \times 6 = 30$ , $15 \times 11 = 165$ , $25 \times 14 = 350$ , $35 \times 9 = 315$ , giving $30 + 165 + 350 + 315 = 860$ .
M1 (dep): the $fx$ products summed; the mark scheme allows one arithmetic slip here
Divide by the total frequency, which is $6 + 11 + 14 + 9 = 40$ , not by the four classes: $860 \div 40 = 21.5$ minutes.
A1: 21.5 minutes, and it lies inside the table's range, which is the check to make
For (c), the median is at position $\frac{40 + 1}{2} = 20.5$ . Running totals down the frequency column are 6, then 17, then 31, so the 20.5th value falls in $20 < t \le 30$ .
B1: the class containing the median, found from a cumulative total rather than by taking the middle row

Practice questions

These are original questions in Edexcel 1MA1 Higher style. There are no diagrams here, so every table is written out as a list of classes with their frequencies, and the frequency polygon questions describe the points rather than showing them. Answers that need rounding say so in the stem and carry a matching tolerance, so work to the stated accuracy rather than truncating early.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Exam pace [2 marks]

The table shows the marks scored by 80 students in a test. $0 < s \le 20$ : 9 students $20 < s \le 40$ : 17 students $40 < s \le 60$ : 24 students $60 < s \le 80$ : 20 students $80 < s \le 100$ : 10 students Which class contains the median mark?

Answer options for question 1
$20 < s \le 40$ $40 < s \le 60$ $60 < s \le 80$ 24
Question 2 Exam pace [2 marks]

A frequency polygon is drawn for a grouped frequency table. One of the classes is $20 < t \le 30$ and its frequency is 14. Write down the coordinates of the point that represents this class.

x = y =
Question 3 Exam pace [3 marks]

The table shows the marks of 40 students in a test marked out of 80. $0 < s \le 20$ : 7 students $20 < s \le 40$ : 12 students $40 < s \le 60$ : 13 students $60 < s \le 80$ : 8 students Work out an estimate for the mean mark.

Your answer (fractions like 1/2 are fine)
Question 4 Stretch [3 marks]

The table shows the masses, in kg, of 25 puppies. $2.5 < w \le 3.0$ : 3 puppies $3.0 < w \le 3.5$ : 7 puppies $3.5 < w \le 4.0$ : 9 puppies $4.0 < w \le 4.5$ : 6 puppies Work out an estimate for the mean mass, in kg.

Your answer (fractions like 1/2 are fine)
Question 5 Stretch [4 marks]

The total frequency of this table is 60. $0 < t \le 10$ : 9 $10 < t \le 20$ : $x$ $20 < t \le 30$ : 21 $30 < t \le 40$ : 12 Work out the value of $x$ , and write down the midpoint of the class that contains the median.

x = midpoint of the median class =
Question 6 Stretch [1 mark]

Two frequency polygons are drawn on the same axes for two groups of 60 people, using the same classes. Polygon P has its highest point at $(15, 22)$ . Polygon Q has its highest point at $(35, 22)$ . Which statement is correct?

Answer options for question 6
Q has a higher modal class than P, so the values in group Q are typically larger The two groups have the same modal class Group P contains more people than group Q Nothing can be compared, because a frequency polygon does not show frequencies

Common mistakes examiners see

Using the end of each class instead of the midpoint, so classes running to 10, 20, 30 and 40 give products of $10 \times 6$ , $20 \times 11$ and so on, and an estimated mean of 26.5 rather than 21.5.

Write the midpoint column before you multiply anything: add the two boundaries and halve. Consistent end values still collect both method marks, so if you notice late, leave the working visible rather than crossing it out.
Dividing the total of the $fx$ column by the number of classes, giving $860 \div 4 = 215$ minutes for a table where nobody took longer than 40.

Total the frequency column and write that number down before dividing. Then compare your answer with the classes: an estimated mean has to land inside the range the table covers, so 215 is a signal to go back, not a result.
Answering the modal class with a frequency, writing 19 instead of $170 < h \le 180$ , or choosing the middle class because it sits in the centre of the table.

Underline the largest number in the frequency column, then copy out the interval on that row. The answer to a modal class question always has an inequality in it, never a bare count.
Giving a single number as the median of grouped data, or taking the middle row of the table as the median class without counting.

The raw values are gone, so the answer is a class. Work out $\frac{n+1}{2}$ , run a cumulative total down the frequency column, and stop on the row where the total first reaches that position.
Plotting a frequency polygon at the class boundaries, or joining the last point back to the first to close the outline.

Each point goes at the midpoint of its class, at a height equal to the frequency, and the points are joined in order with straight line segments and left open. Closing the shape or plotting at the ends loses the plotting marks even when the table was right.

Frequently asked questions

Why is the mean from a grouped frequency table only an estimate?: Because the table records which class each value fell into and nothing more, so the exact total of the data cannot be recovered. Using the midpoint of a class assumes every value in it sits at the centre, which is almost never true. That assumption is what makes the answer an estimate, and it is why Edexcel writes the instruction as work out an estimate for the mean rather than calculate the mean.
Are grouped frequency tables on the Foundation paper?: Yes. The estimated mean, the modal class and the class containing the median are all crossover content on Edexcel 1MA1, so a Foundation and a Higher paper in the same series can carry very similar questions. The Higher version tends to add a second demand on top, such as a missing frequency to find or a comparison of two distributions to write.
Where do the points go on a frequency polygon?: At the midpoint of each class, at a height equal to the frequency of that class, joined in order with straight line segments. A class of $20 < t \le 30$ with frequency 14 is plotted at 25 across and 14 up. Points plotted at the ends of the intervals, and outlines closed back to the first point, are the two errors examiners report on this part.
How many marks is an estimated mean question worth on Edexcel Higher?: Three marks in the standard form: one for the $fx$ products, one dependent mark for dividing by the total frequency, and one for the answer. With a modal class part and a frequency polygon or a comparison alongside it, the whole question usually comes to 3-5 marks.

Grouped Frequency Tables - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics