Sec 2 Maths · Statistics (Sec 2)

Grouped data questions with answers

Practice questions on grouped data, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

The table shows the distances some members of a cycling club rode on Sunday. Write down the modal class.

Distance (d km)	Frequency
$0 < d \le 20$	4
$20 < d \le 40$	2
$40 < d \le 60$	13
$60 < d \le 80$	11

Step 1 of 2

Find the highest frequency

The highest frequency is

13

Try a question: Grouped data

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Questions

Question 1The modal class
The table shows the heights of some plants. Write down the modal class.
Height (h cm) Frequency
$10 < h \le 20$ 3
$20 < h \le 30$ 14
$30 < h \le 40$ 5
$40 < h \le 50$ 9
Show the worked solution
1. Step 1: Find the highest frequencyThe highest frequency is $14$ .
2. Step 2: The modal class is the class with the highest frequencyThe modal class is $20 < h \le 30$ .
Question 2The modal class
The table shows the distances some members of a cycling club rode on Sunday. Write down the modal class.
Distance (d km) Frequency
$20 < d \le 40$ 10
$40 < d \le 60$ 10
$60 < d \le 80$ 3
$80 < d \le 100$ 12
$100 < d \le 120$ 8
Show the worked solution
1. Step 1: Find the highest frequencyThe highest frequency is $12$ .
2. Step 2: The modal class is the class with the highest frequencyThe modal class is $80 < d \le 100$ .
Question 3Estimate the mean from mid-values
The table shows the heights of some plants. Using the mid-value of each class, calculate an estimate of the mean, correct to 3 significant figures.
Height (h cm) Frequency
$20 < h \le 30$ 12
$30 < h \le 40$ 7
$40 < h \le 50$ 4
$50 < h \le 60$ 15
$60 < h \le 70$ 11
Show the worked solution
1. Step 1: Mid-value of each class: halfway between its endsMid-values: $25,\ 35,\ 45,\ 55,\ 65$
2. Step 2: Total frequencyTotal frequency $= 12 + 7 + 4 + 15 + 11$ $= 49$
3. Step 3: Estimated total: each mid-value × its frequency, addedTotal $= 25 \times 12 + 35 \times 7 + 45 \times 4 + 55 \times 15 + 65 \times 11$ $= 2265$
4. Step 4: Estimated mean = total ÷ total frequencyMean $= \dfrac{2265}{49}$ $= 46.2244\ldots$ $= 46.2$ cm (3 s.f.)
Question 4Estimate the mean from mid-values
The table shows the marks of some pupils in a test. Using the mid-value of each class, calculate an estimate of the mean.
Marks (x) Frequency
$20 < x \le 24$ 8
$24 < x \le 28$ 2
$28 < x \le 32$ 9
$32 < x \le 36$ 7
$36 < x \le 40$ 14
Show the worked solution
1. Step 1: Mid-value of each class: halfway between its endsMid-values: $22,\ 26,\ 30,\ 34,\ 38$
2. Step 2: Total frequencyTotal frequency $= 8 + 2 + 9 + 7 + 14$ $= 40$
3. Step 3: Estimated total: each mid-value × its frequency, addedTotal $= 22 \times 8 + 26 \times 2 + 30 \times 9 + 34 \times 7 + 38 \times 14$ $= 1268$
4. Step 4: Estimated mean = total ÷ total frequencyMean $= \dfrac{1268}{40}$ $= 31.7$ marks

Question 5The class containing the median

The table shows the distances some members of a cycling club rode on Sunday. Find the class that contains the median.

Distance (d km)	Frequency
$20 < d \le 40$	6
$40 < d \le 60$	3
$60 < d \le 80$	13
$80 < d \le 100$	10
$100 < d \le 120$	5
$120 < d \le 140$	14

Show the worked solution

Step 1: Total frequency: where is the middle?Total frequency $= 6 + 3 + 13 + 10 + 5 + 14$ $= 51$ The median is the $\dfrac{51 + 1}{2} = 26$ th value.
Step 2: Running totals of the frequencies, from the first classRunning totals: $6,\ 9,\ 22,\ 32,\ 37,\ 51$
Step 3: The median is in the first class whose running total reaches that positionThe running total first reaches $26$ in the class $80 < d \le 100$ , so the median lies in it.

Question 6The class containing the median
The table shows the time taken by some pupils to finish a puzzle. Find the class that contains the median.
Time (t minutes) Frequency
$0 < t \le 5$ 5
$5 < t \le 10$ 9
$10 < t \le 15$ 9
$15 < t \le 20$ 15
$20 < t \le 25$ 2
Show the worked solution
1. Step 1: Total frequency: where is the middle?Total frequency $= 5 + 9 + 9 + 15 + 2$ $= 40$ The median is the mean of the $20$ th and $21$ th values.
2. Step 2: Running totals of the frequencies, from the first classRunning totals: $5,\ 14,\ 23,\ 38,\ 40$
3. Step 3: The median is in the first class whose running total reaches that positionThe running total first reaches $20$ in the class $10 < t \le 15$ , so the median lies in it.

More on statistics (sec 2)

Notes: Statistics (Sec 2)Worksheet with answers (PDF)

Endless questions on grouped data

A free account gives you new questions every time, step-by-step help, and tracks what you've mastered.

Start free

Worked example

The table shows the distances some members of a cycling club rode on Sunday. Write down the modal class.

Distance (d km)	Frequency
$0 < d \le 20$	4
$20 < d \le 40$	2
$40 < d \le 60$	13
$60 < d \le 80$	11

Step 1 of 2

Find the highest frequency

The highest frequency is

13

Questions

Question 1The modal class

The table shows the heights of some plants. Write down the modal class.

Height (h cm)	Frequency
$10 < h \le 20$	3
$20 < h \le 30$	14
$30 < h \le 40$	5
$40 < h \le 50$	9

Show the worked solution

Step 1: Find the highest frequencyThe highest frequency is $14$ .
Step 2: The modal class is the class with the highest frequencyThe modal class is $20 < h \le 30$ .

Question 2The modal class

The table shows the distances some members of a cycling club rode on Sunday. Write down the modal class.

Distance (d km)	Frequency
$20 < d \le 40$	10
$40 < d \le 60$	10
$60 < d \le 80$	3
$80 < d \le 100$	12
$100 < d \le 120$	8

Show the worked solution

Step 1: Find the highest frequencyThe highest frequency is $12$ .
Step 2: The modal class is the class with the highest frequencyThe modal class is $80 < d \le 100$ .

Question 3Estimate the mean from mid-values

The table shows the heights of some plants. Using the mid-value of each class, calculate an estimate of the mean, correct to 3 significant figures.

Height (h cm)	Frequency
$20 < h \le 30$	12
$30 < h \le 40$	7
$40 < h \le 50$	4
$50 < h \le 60$	15
$60 < h \le 70$	11

Show the worked solution

Step 1: Mid-value of each class: halfway between its endsMid-values: $25,\ 35,\ 45,\ 55,\ 65$
Step 2: Total frequencyTotal frequency $= 12 + 7 + 4 + 15 + 11$ $= 49$
Step 3: Estimated total: each mid-value × its frequency, addedTotal $= 25 \times 12 + 35 \times 7 + 45 \times 4 + 55 \times 15 + 65 \times 11$ $= 2265$
Step 4: Estimated mean = total ÷ total frequencyMean $= \dfrac{2265}{49}$ $= 46.2244\ldots$ $= 46.2$ cm (3 s.f.)

Question 4Estimate the mean from mid-values

The table shows the marks of some pupils in a test. Using the mid-value of each class, calculate an estimate of the mean.

Marks (x)	Frequency
$20 < x \le 24$	8
$24 < x \le 28$	2
$28 < x \le 32$	9
$32 < x \le 36$	7
$36 < x \le 40$	14

Show the worked solution

Step 1: Mid-value of each class: halfway between its endsMid-values: $22,\ 26,\ 30,\ 34,\ 38$
Step 2: Total frequencyTotal frequency $= 8 + 2 + 9 + 7 + 14$ $= 40$
Step 3: Estimated total: each mid-value × its frequency, addedTotal $= 22 \times 8 + 26 \times 2 + 30 \times 9 + 34 \times 7 + 38 \times 14$ $= 1268$
Step 4: Estimated mean = total ÷ total frequencyMean $= \dfrac{1268}{40}$ $= 31.7$ marks

Question 5The class containing the median

The table shows the distances some members of a cycling club rode on Sunday. Find the class that contains the median.

Distance (d km)	Frequency
$20 < d \le 40$	6
$40 < d \le 60$	3
$60 < d \le 80$	13
$80 < d \le 100$	10
$100 < d \le 120$	5
$120 < d \le 140$	14

Show the worked solution

Step 1: Total frequency: where is the middle?Total frequency $= 6 + 3 + 13 + 10 + 5 + 14$ $= 51$ The median is the $\dfrac{51 + 1}{2} = 26$ th value.
Step 2: Running totals of the frequencies, from the first classRunning totals: $6,\ 9,\ 22,\ 32,\ 37,\ 51$
Step 3: The median is in the first class whose running total reaches that positionThe running total first reaches $26$ in the class $80 < d \le 100$ , so the median lies in it.

Question 6The class containing the median

The table shows the time taken by some pupils to finish a puzzle. Find the class that contains the median.

Time (t minutes)	Frequency
$0 < t \le 5$	5
$5 < t \le 10$	9
$10 < t \le 15$	9
$15 < t \le 20$	15
$20 < t \le 25$	2

Show the worked solution

Step 1: Total frequency: where is the middle?Total frequency $= 5 + 9 + 9 + 15 + 2$ $= 40$ The median is the mean of the $20$ th and $21$ th values.
Step 2: Running totals of the frequencies, from the first classRunning totals: $5,\ 14,\ 23,\ 38,\ 40$
Step 3: The median is in the first class whose running total reaches that positionThe running total first reaches $20$ in the class $10 < t \le 15$ , so the median lies in it.