Find the mean and median of the following data: Class 85-90 90-95 95-100 100-105 105-110 110-115 --- --- --- --- --- --- --- Frequency 15 22 20 18 20 25

MathsCBSE Class 10 · Standard · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2024-25: question 35(A)

From the official CBSE sample question paper for 2024-25, section D, question 35(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(A) · 5 marks · Long answerStatistics

Find the mean and median of the following data:

Class	85-90	90-95	95-100	100-105	105-110	110-115
Frequency	15	22	20	18	20	25

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

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

Hint 1

Find the class marks, then use the step-deviation method with $a = 102.5$ and $h = 5$ .

Hint 2

For the median, add a cumulative frequency column and locate the class containing the $6 0^{t h}$ observation.

Hint 3

Use median $= l + \frac{N /2 - c f}{f} \times h$ with the values for that class.

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 the assumed mean $a = 102.5$ and class size $h = 5$ , with $u_{i} = \frac{x _{i} - 102.5}{5}$ .
Class $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$ c.f.
85-90 87.5 15 -3 -45 15
90-95 92.5 22 -2 -44 37
95-100 97.5 20 -1 -20 57
100-105 102.5 18 0 0 75
105-110 107.5 20 1 20 95
110-115 112.5 25 2 50 120
Total 120 -39
2 marks
Mean $\overset{x}{ˉ} = a + h \cdot \frac{\sum f _{i} u _{i}}{\sum f _{i}}$
$= 102.5 + 5 \times \frac{- 39}{120}$
$= 102.5 - 1.625 = 100.875$
1 mark
$N = 120$ , so $\frac{N}{2} = 60$ .
½ mark
The first cumulative frequency greater than 60 is 75, so the median class is $100$ – $105$ .
Here $l = 100$ , $c f = 57$ , $f = 18$ , $h = 5$ .
½ mark
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h$
$= 100 + \frac{60 - 57}{18} \times 5$
$= 100 + 0.83 = 100.83$ (approx.)
1 mark

Class	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$	c.f.
85-90	87.5	15	-3	-45	15
90-95	92.5	22	-2	-44	37
95-100	97.5	20	-1	-20	57
100-105	102.5	18	0	0	75
105-110	107.5	20	1	20	95
110-115	112.5	25	2	50	120
Total		120		-39

Answer: Mean = 100.875; Median ≈ 100.83

Where the marks are

Draws a correct table with class marks, $u_{i}$ , $f_{i} u_{i}$ and cumulative frequencies (2 marks)
Applies the step-deviation formula to get mean $= 100.875$ (1 mark)
Finds $N /2 = 60$ (½ mark)
Identifies the median class 100–105 (½ mark)
Applies the median formula to get $\approx 100.83$ (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 paperSample paper 2024-25 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2024-25: question 35(A)

Hints

Step-by-step solution

Revise the chapter

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