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Maharashtra State BoardSSC (English Medium) 10th Standard

The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices. Price (Rupees) 20 less than, 20 - 40, 40 - 60, 60 - 80 - Algebra Mathematics 1

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Questions

The prices of different articles and demand for them is shown in the following frequency distribution table. Find the median of the prices.

Price (Rupees)
20 less than 20 - 40 40 - 60 60 - 80 80 - 100
No. of articles 140 100 80 60 20

In a general store, the prices of different articles and their demand are shown in the following frequency distribution table. Find the median of the prices.

Prices in rupees No. of articles
Less than 20 140
20-40 100
40-60 80
60-80 60
80-100 20
Sum
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Solution

Class
(Prices in rupees)
Frequency
(Number of articles)
(fi)
Cumulative frequency
less than the
upper limit 
20 less than 140 140 → cf
20 - 40 100 → f 240
40 - 60 80 320
60 - 80 60 380
80 - 100 20 400
Total N = 400  

From the above table, we get

∑fi = N = 400

∴ `N/2 = 400/2 = 200`

∴ The cumulative frequency greater than (or equal to) 200 is 240.

∴ 20 – 40 is the median class.

Now, L = 20, f = 100, c.f. = 140, h = 20

Now, Median = `L + [(N/2 - c.f.)/f] xx h`

= `20+ ((200-140)/100) xx20`

= `20 + (60/100) xx 20`

= 20 + 12

= ₹ 32

Hence, the median of the prices of an article is ₹ 32.

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Chapter 6: Statistics - Miscellaneous Problems 6 [Page 166]

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Balbharati Algebra (Mathematics 1) [English] Standard 10 Maharashtra State Board
Chapter 6 Statistics
Miscellaneous Problems 6 | Q 7 | Page 166
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