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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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