Consider the following distribution of hourly wages of 50 workers of a factory: Hourly wages (in ₹) 100-120 120-140 140-160 160-180 180-200 --- --- --- --- --- --- Number of workers 12 14 8 6 10 Find the mean and the median of the above data.

MathsCBSE Class 10 · Basic · Board exam 2024, set 430/3/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2024: question 32

Set in the CBSE Class 10 board exam of 2024, paper 430/3/1, section D, question 32, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32 · 5 marks · Long answerStatistics

Consider the following distribution of hourly wages of $50$ workers of a factory:

Hourly wages (in ₹)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Find the mean and the median of the above data.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/3/1.

Hints

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

Hint 1

Find class marks and use step-deviation with assumed mean 150 and $h = 20$ .

Hint 2

For the median, locate the class where cumulative frequency first reaches $N /2 = 25$ .

Hint 3

Median $= l + \frac{N /2 - c f}{f} \times h$ , where cf is the cumulative frequency of the class before the median 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 assumed mean $a = 150$ , class size $h = 20$ , $u_{i} = \frac{x _{i} - 150}{20}$ .
C.I. $x_{i}$ $f_{i}$ $u_{i}$ $f_{i} u_{i}$ c.f.
100-120 110 12 −2 −24 12
120-140 130 14 −1 −14 26
140-160 150 8 0 0 34
160-180 170 6 1 6 40
180-200 190 10 2 20 50
Total 50 −12
2½ marks
Mean $= a + \frac{\sum f _{i} u _{i}}{\sum f _{i}} \times h = 150 + \frac{- 12}{50} \times 20 = 150 - 4.8 = 145.2$ . Mean wage = ₹145.20
1½ marks
$\frac{N}{2} = 25$ ; median class is 120-140 with $l = 120$ , $c f = 12$ , $f = 14$ , $h = 20$ .
Median $= l + \frac{\frac{N}{2} - c f}{f} \times h = 120 + \frac{25 - 12}{14} \times 20 = 120 + 18.57 = 138.57$ (approx.). Median wage ≈ ₹138.57
1 mark

C.I.	$x_{i}$	$f_{i}$	$u_{i}$	$f_{i} u_{i}$	c.f.
100-120	110	12	−2	−24	12
120-140	130	14	−1	−14	26
140-160	150	8	0	0	34
160-180	170	6	1	6	40
180-200	190	10	2	20	50
Total		50		−12

Answer: Mean = ₹145.2; Median ≈ ₹138.57

Where the marks are

Correct table with class marks, $u_{i}$ , $f_{i} u_{i}$ and cumulative frequencies (2½ marks)
Mean by step-deviation $= 145.2$ (1½ marks)
Median using formula $= 138.57$ (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 paper2024 board paper, set 430/3/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 32

Hints

Step-by-step solution

Revise the chapter

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