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प्रश्न
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 |
उत्तर
|
Class
(Prices in Rupees) |
Frequency (Number of articles) (fi) |
Cumulaive 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 |
| 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, cf = 140, h = 20
Now, Median = `L + [(N/2 - cf)/f] xx h`
`= 20+ ((400/2-140)/100) xx20`
`= 20 + ((200 - 140)/100) xx 20`
= `20 + (60/100) xx 20`
= 20 + 12
= Rs. 32
Hence, the median of the prices is Rs. 32.
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| 0 - 10 | 5 |
| 10 - 20 | x |
| 20 - 30 | 20 |
| 30 - 40 | 15 |
| 40 - 50 | y |
| 50 - 60 | 5 |
| Total | 60 |
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| Marks | No. of students |
| Less than 10 | 0 |
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| Less than 50 | 25 |
| Less than 70 | 43 |
| Less than 90 | 65 |
| Less than 110 | 87 |
| Less than 130 | 96 |
| Less than 150 | 100 |
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|
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|
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