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Question
An incomplete distribution is given below:
| Variable: | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
| Frequency: | 12 | 30 | - | 65 | - | 25 | 18 |
You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.
Solution
(i)
| Class interval | Frequency | Cumulative frequency |
| 10-20 | 12 | 12 |
| 20-30 | 30 | 42 |
| 30-40 | x | 42 + x |
| 40-50 | 65 | 107 + x |
| 50-60 | y | 107 + x + y |
| 60-70 | 25 | 132 + x + y |
| 70-80 | 18 | 150 + x + y |
| N = 230 |
Given median = 46
Then, median class = 40 - 50
l = 40, h = 50 - 40 = 10, f = 65, F = 42 + x
Median `=l+((N/2)-F)/fxxh`
`rArr46=40+(115-(42+x))/65xx10`
`rArr46 - 40 = (115-42-x)/65xx10`
`rArr6=(73-x)/65=10`
`rArr(6xx65)/10=73-x`
`rArr390/10=73-x`
39 = 73 - x
x = 73 - 39
x = 34
Given N = 230
⇒ 12 + 30 + x + 65 + y + 25 + 18 = 230
⇒ 12 + 30 + 34 + 65 + y + 25 + 18 = 230
⇒ 184 + y = 230
⇒ y = 230 - 184
⇒ y = 46
(ii)
| Class interval | Mid value(x) | Frequency(f) | fx |
| 10-20 | 15 | 12 | 180 |
| 20-30 | 25 | 30 | 750 |
| 30-40 | 35 | 34 | 1190 |
| 40-50 | 45 | 65 | 2925 |
| 50-60 | 55 | 46 | 2530 |
| 60-70 | 65 | 25 | 1625 |
| 70-80 | 75 | 18 | 1350 |
| N = 230 | `sumfx=10550` |
Mean `=(sumfx)/N`
`=10550/230=45.87`
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