1. Find missing frequency from the following data
| Class | Frequency |
| 40 - 59 | 50 |
| 60 - 79 | ? |
| 80 - 99 | 500 |
| 100 - 119 | ? |
| 120 - 139 | 50 |
Total Frequency (N) = 1000 and median = 87.50
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | `cf`
`(3)` |
| 40-59 | 50 | 50 `50=0+50`
`(3)=`Previous `(3)+(2)` |
| 60-79 | a | 50 + a `50 + a=50+a`
`(3)=`Previous `(3)+(2)` |
| 80-99 | 500 | 550 + a `550 + a=50 + a+500`
`(3)=`Previous `(3)+(2)` |
| 100-119 | b | 550 + a + b `550 + a + b=550 + a+b`
`(3)=`Previous `(3)+(2)` |
| 120-139 | 50 | 600 + a + b `600 + a + b=550 + a + b+50`
`(3)=`Previous `(3)+(2)` |
| --- | --- | --- |
| -- | `n=1000`
`n=a + b + 600` | |
`n=1000`
`a+b+600=1000`
`a+b=400 ->(1)`
To find median class
Here, median is `87.5`.
`:.` The median class is `79.5 - 99.5`.
Now,
`:. L = `lower boundary point of median class `=79.5`
`:. n = `Total frequency `=1000`
`:. cf = `Cumulative frequency of the class preceding the median class `=50 + a`
`:. f = `Frequency of the median class `=500`
`:. c = `class length of median class `=20`
Median `M = L + (( n)/2 - cf)/f * c`
`87.5=79.5 + (( 1000)/2 - (50 + a))/500 * 20`
`87.5 - 79.5=(500 - (50 + a))/500 * 20`
`8 = (-a+450)/500 * 20`
`8*500=(-a+450)*20`
`4000=-20a+9000`
`20a=5000`
`a=5000/20`
`a=250`
Substituting in `(1)`
`250 + b = 400`
`b = 400 - 250`
`b = 150`
Thus, the missing frequencies are `250` and `150` respectively.
2. Find missing frequency from the following data
| Class | Frequency |
| 11 - 20 | 42 |
| 21 - 30 | 38 |
| 31 - 40 | a |
| 41 - 50 | 54 |
| 51 - 60 | b |
| 61 - 70 | 36 |
| 71 - 80 | 32 |
Total Frequency (N) = 400 and median = 38.5
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | `cf`
`(6)` |
| 11 - 20 | 42 | 42 |
| 21 - 30 | 38 | 80 |
| 31 - 40 | a | 80+a |
| 41 - 50 | 54 | 134+a |
| 51 - 60 | b | 134+a+b |
| 61 - 70 | 36 | 170+a+b |
| 71 - 80 | 32 | 202+a+b |
| --- | --- | |
| `n=202+a+b` | |
`n = 400`
`202 + a+b= 400`
`a+b=198 ->(1)`
To find median class
Here, median is `38.5`.
`:.` The median class is `30.5 - 40.5`.
Now,
`:. L = `lower boundary point of median class `=30.5`
`:. n = `Total frequency `=400`
`:. cf = `Cumulative frequency of the class preceding the median class `=80`
`:. f = `Frequency of the median class `=a`
`:. c = `class length of median class `=10`
Median `M = L + (( n)/2 - cf)/f * c`
`38.5=30.5 + (( 400)/2 - (80))/a * 10`
`38.5 - 30.5=(200 - (80))/a * 10`
`8=(120)/a * 10`
`8*a=(120)*10`
`8a=1200`
`8a-1200 = 0`
`8a = 1200`
`a = 1200/8`
`a = 150 ->(2)`
Now solving this 2 equations using substitution method
`a+b=198`
and `a=150`
`a+b=198 ->(1)`
`a=150 ->(2)`
Putting `a=150` in equation `(1)`, we get
`=>a+b=198`
`=>150+b=198`
`=>b=198-150`
`=>b=48`
`:.a=150" and "b=48`
Thus, the missing frequencies are `150 and 48` respectively.
This material is intended as a summary. Use your textbook for detail explanation.
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