Medical check-up was carried out for 35 students of a class and their weights were recorded as follows: Weight (in kg) 38-40 40-42 42-44 44-46 46-48 48-50 50-52 --- --- --- --- --- --- --- --- Number of Students 3 2 4 5 14 4 3 Find the difference between the mean weight and the median weight.

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2025: question 35

Set in the CBSE Class 10 board exam of 2025, paper 30/5/1, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerStatistics

Medical check-up was carried out for 35 students of a class and their weights were recorded as follows:

Weight (in kg)	38-40	40-42	42-44	44-46	46-48	48-50	50-52
Number of Students	3	2	4	5	14	4	3

Find the difference between the mean weight and the median weight.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/5/1.

Hints

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

Hint 1

You need two separate averages here: the mean (from $\sum f_{i} x_{i}$ or step deviation) and the median (from cumulative frequencies).

Hint 2

For the mean, take $a = 45$ and $h = 2$ to keep the numbers small. For the median, find $n /2$ and locate the class whose cumulative frequency first exceeds it.

Hint 3

Use Median $= l + \frac{n /2 - c f}{f} \times h$ , then subtract the smaller average from the larger one.

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 $a = 45$ and class width $h = 2$ , so $u_{i} = \frac{x _{i} - 45}{2}$ .
Weight (kg) $f_{i}$ $x_{i}$ $u_{i}$ $f_{i} u_{i}$ c.f.
38–40 3 39 −3 −9 3
40–42 2 41 −2 −4 5
42–44 4 43 −1 −4 9
44–46 5 45 0 0 14
46–48 14 47 1 14 28
48–50 4 49 2 8 32
50–52 3 51 3 9 35
Total 35 14
1½ marks
Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 45 + \frac{14}{35} \times 2$
1 mark
$= 45 + 0.8 = 45.8$ kg
½ mark
$\frac{n}{2} = \frac{35}{2} = 17.5$ . The first c.f. greater than 17.5 is 28, so the median class is $46$ – $48$ , with $l = 46$ , $c f = 14$ , $f = 14$ , $h = 2$ .
½ mark
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 46 + \frac{17.5 - 14}{14} \times 2$
½ mark
$= 46 + \frac{3.5}{14} \times 2 = 46 + 0.5 = 46.5$ kg
½ mark
Difference between the median and the mean $= 46.5 - 45.8 = 0.7$ kg
½ mark

Weight (kg)	$f_{i}$	$x_{i}$	$u_{i}$	$f_{i} u_{i}$	c.f.
38–40	3	39	−3	−9	3
40–42	2	41	−2	−4	5
42–44	4	43	−1	−4	9
44–46	5	45	0	0	14
46–48	14	47	1	14	28
48–50	4	49	2	8	32
50–52	3	51	3	9	35
Total	35			14

Answer: Mean = 45.8 kg, Median = 46.5 kg, Difference = 0.7 kg

Where the marks are

Correct table with class marks, $u_{i}$ , $f_{i} u_{i}$ and cumulative frequencies (1½ marks)
Correct substitution in the mean formula (1 mark)
Mean = 45.8 kg (½ mark)
Identifies the median class as 46–48 (½ mark)
Correct substitution in the median formula (½ mark)
Median = 46.5 kg (½ mark)
Difference = 0.7 kg (½ 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 paper2025 board paper, set 30/5/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2025: question 35

Hints

Step-by-step solution

Revise the chapter

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