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Question
The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data:
| Class interval | Frequency |
| 0 - 100 | 2 |
| 100 - 200 | 5 |
| 200 - 300 | f1 |
| 300 - 400 | 12 |
| 400 - 500 | 17 |
| 500 - 600 | 20 |
| 600 - 700 | f2 |
| 700 - 800 | 9 |
| 800 - 900 | 7 |
| 900 - 1000 | 4 |
Solution
| Class interval | Frequency | Cumulative frequency |
| 0 - 100 | 2 | 2 |
| 100 - 200 | 5 | 7 |
| 200 - 300 | f1 | 7 + f1 |
| 300 - 400 | 12 | 19 + f1 |
| 400 - 500 | 17 | 36 + f1 |
| 500 - 600 | 20 | 56 + f1 |
| 600 - 700 | f2 | 56 + f1 + f2 |
| 700 - 800 | 9 | 65 + f1 + f2 |
| 800 - 900 | 7 | 72 + f1 + f2 |
| 900 - 1000 | 4 | 76 + f1 + f2 |
| N = 100 |
Given median = 525
Then median class = 500 - 600
l = 500, f = 20, F = 36 + f1 and h = 600 - 500 = 100
Median `l+(N/2-F)/fxxh`
`rArr525=500+(50-(36+f1))/20xx100`
`rArr525-500=(50-36-f1)/20xx100`
⇒ 25 = (14 - f1) x 5
⇒ 25 = 70 - 5f1
⇒ 5f1 = 70 - 25
⇒ 5f1 = 45
⇒ f1 = 45/5 = 9
Given sum of frequencies = 100
⇒ 2 + 5 + f1 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100
⇒ 2 + 5 + 9 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100
⇒ 85 + f2 = 100
⇒ f2 = 100 - 85
⇒ f2 = 15
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