Find the mean and the median of the following data: Marks Number of Students --- --- 0 – 10 3 10 – 20 5 20 – 30 16 30 – 40 12 40 – 50 13 50 – 60 20 60 – 70 6 70 – 80 5

MathsCBSE Class 10 · Basic · Board exam 2023, set 430/2/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2023: question 33

Set in the CBSE Class 10 board exam of 2023, paper 430/2/1, section D, question 33, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33 · 5 marks · Long answerStatistics

Find the mean and the median of the following data:

Marks	Number of Students
0 – 10	3
10 – 20	5
20 – 30	16
30 – 40	12
40 – 50	13
50 – 60	20
60 – 70	6
70 – 80	5

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/2/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Find the class marks and use the step-deviation method with assumed mean 35.

Hint 2

For the median, add a cumulative frequency column and locate the class containing the 40th observation.

Hint 3

Use median $= l + \frac{n /2 - c f}{f} \times h$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Take assumed mean $a = 35$ , class size $h = 10$ , $u_{i} = \frac{x _{i} - 35}{10}$ .
Marks $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$ cf
0–10 5 3 −3 −9 3
10–20 15 5 −2 −10 8
20–30 25 16 −1 −16 24
30–40 35 12 0 0 36
40–50 45 13 1 13 49
50–60 55 20 2 40 69
60–70 65 6 3 18 75
70–80 75 5 4 20 80
Total 80 56
2 marks
Mean $= a + h \cdot \frac{\sum f _{i} u _{i}}{\sum f _{i}} = 35 + 10 \times \frac{56}{80} = 35 + 7 = 42$
1½ marks
$\frac{n}{2} = \frac{80}{2} = 40$ . The cumulative frequency just greater than 40 is 49, so the median class is 40–50, with $l = 40$ , $c f = 36$ , $f = 13$ , $h = 10$ .
½ mark
Median $= l + (\frac{\frac{n}{2} - c f}{f}) \times h = 40 + \frac{40 - 36}{13} \times 10 = 40 + \frac{40}{13} \approx 40 + 3.08 = 43.08$ (≈ 43.1)
1 mark

Marks	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$	cf
0–10	5	3	−3	−9	3
10–20	15	5	−2	−10	8
20–30	25	16	−1	−16	24
30–40	35	12	0	0	36
40–50	45	13	1	13	49
50–60	55	20	2	40	69
60–70	65	6	3	18	75
70–80	75	5	4	20	80
Total		80		56

Answer: Mean = 42; Median ≈ 43.08 (about 43.1)

Where the marks are

Correct table with class marks, $f_{i} u_{i}$ and cumulative frequencies (2 marks)
Mean $= 35 + 10 \times \frac{56}{80} = 42$ (1½ marks)
Identifies median class 40–50 (1/2 mark)
Median $= 40 + \frac{4}{13} \times 10 \approx 43.1$ (1 mark)

Revise the chapter

Chapter: Statistics (notes, formulas, where marks are lost and every past question)
Topic: Mean of Grouped Data: Direct Method

Practise this chapterBoard-style questions from Statistics, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 430/2/1 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic 2023: question 33

Hints

Step-by-step solution

Revise the chapter

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