The monthly expenditure on milk in 200 families of a Housing Society is given below: Monthly Expenditure (in ₹) 1000-1500 1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000 --- --- --- --- --- --- --- --- --- Number of families 24 40 33 x 30 22 16 7 Find the value of x and also, find the median and mean expenditure on milk.

MathsCBSE Class 10 · Standard · Board exam 2023, set 30/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2023: question 35

Set in the CBSE Class 10 board exam of 2023, paper 30/4/1, 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

The monthly expenditure on milk in 200 families of a Housing Society is given below:

Monthly Expenditure (in ₹)	1000-1500	1500-2000	2000-2500	2500-3000	3000-3500	3500-4000	4000-4500	4500-5000
Number of families	24	40	33	x	30	22	16	7

Find the value of x and also, find the median and mean expenditure on milk.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/4/1.

Hints

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

Hint 1

The frequencies must add up to 200 families.

Hint 2

For the median, find the class where the cumulative frequency first passes $N /2 = 100$ .

Hint 3

For the mean, use step deviation with $a = 2750$ and $h = 500$ .

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 $a = 2750$ , class width $h = 500$ and $u_{i} = \frac{x _{i} - 2750}{500}$ .

Class	$x_{i}$	$f_{i}$	c.f.	$u_{i}$	$f_{i} u_{i}$
1000–1500	1250	24	24	−3	−72
1500–2000	1750	40	64	−2	−80
2000–2500	2250	33	97	−1	−33
2500–3000	2750	28	125	0	0
3000–3500	3250	30	155	1	30
3500–4000	3750	22	177	2	44
4000–4500	4250	16	193	3	48
4500–5000	4750	7	200	4	28
Total		200			−35

2 marks

Sum of frequencies: $24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 172 + x$ .
$172 + x = 200$
$\Rightarrow x = 28$
1 mark
Median: $\frac{N}{2} = 100$ . The c.f. just above 100 is 125, so the median class is 2500–3000.
Here $l = 2500$ , $c f = 97$ , $f = 28$ , $h = 500$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 2500 + \frac{100 - 97}{28} \times 500 = 2500 + 53.57 \approx ₹ 2553.6$
1 mark
Mean: Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 2750 + \frac{- 35}{200} \times 500 = 2750 - 87.5 = ₹ 2662.5$
1 mark

Answer: x = 28; median ≈ ₹2553.6; mean = ₹2662.5

Where the marks are

Correct table with class marks, cumulative frequencies and $f_{i} u_{i}$ (2 marks)
Finds $x = 28$ from total 200 (1 mark)
Identifies median class 2500–3000 and gets median ≈ ₹2553.6 (1 mark)
Gets mean = ₹2662.5 (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 30/4/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2023: question 35

Hints

Step-by-step solution

Revise the chapter

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