5. Find missing frequency from the following data
| Class | Frequency |
| 0 - 10 | 4 |
| 10 - 20 | 16 |
| 20 - 30 | ? |
| 30 - 40 | ? |
| 40 - 50 | ? |
| 50 - 60 | 6 |
| 60 - 70 | 4 |
Total Frequency (N) = 230 and median;mode = 33.5,34
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` | `cf`
`(6)` |
| 0 - 10 | 4 | 4 |
| 10 - 20 | 16 | 20 |
| 20 - 30 | a` = f_0` | 20 + a |
| 30 - 40 | b` = f_1` | 20 + a + b |
| 40 - 50 | c` = f_2` | 20 + a + b + c |
| 50 - 60 | 6 | 26 + a + b + c |
| 60 - 70 | 4 | 30 + a + b + c |
| --- | --- | |
| `n=30+a+b+c` | |
`n = 230`
`30 + a+b+c= 230`
`a+b+c=200 ->(1)`
To find Mode Class
Here, mode is `34`.
`:.` The mode class is `30 - 40`.
`:. L = `lower boundary point of mode class `=30`
`:. f_1 = ` frequency of the mode class `=b`
`:. f_0 = ` frequency of the preceding class `=a`
`:. f_2 = ` frequency of the succedding class `=c`
`:. c = ` class length of mode class `=10`
`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
`34 = 30 + ((b - a)/(2*b - a - c)) * 10`
`34 - 30 = ((b - a)/(2*b - a - c)) * 10`
`4 = ((b - a)/(2*b - a - c)) * 10`
`4 = ((b - a)/(-a+2b-c)) * 10`
`4*(-a+2b-c)=(b-a)*10`
`-4a+8b-4c=-10a+10b`
`6a-2b-4c=0`
`2(3a-b-2c)=2*0`
`3a-b-2c=0 ->(2)`
To find median class
Here, median is `33.5`.
`:.` The median class is `30 - 40`.
Now,
`:. L = `lower boundary point of median class `=30`
`:. n = `Total frequency `=230`
`:. cf = `Cumulative frequency of the class preceding the median class `=20 + a`
`:. f = `Frequency of the median class `=b`
`:. c = `class length of median class `=10`
Median `M = L + (( n)/2 - cf)/f * c`
`33.5=30 + (( 230)/2 - (20 + a))/b * 10`
`33.5 - 30=(115 - (20 + a))/b * 10`
`3.5=(-a+95)/b * 10`
`3.5*b=(-a+95)*10`
`3.5b=-10a+950`
`10a+3.5b=950 ->(3)`
Now solving this 3 equations using substitution method
Total Equations are `3`
`a+b+c=200 -> (1)`
`3a-b-2c=0 -> (2)`
`10a+3.5b+0c=950 -> (3)`
Select the equations `(1)` and `(2)`, and eliminate the variable `c`.
| `a+b+c=200` | ` xx 2->` | | `` | `2a` | `+` | `2b` | `+` | `2c` | `=` | `400` | `` |
| | + | |
| `3a-b-2c=0` | ` xx 1->` | | `` | `3a` | `-` | `b` | `-` | `2c` | `=` | `0` | `` |
| | |
|
| | | `` | `5a` | `+` | `b` | | | `=` | `400` | ` -> (4)` |
Select the equations `(3)` and `(4)`, and eliminate the variable `b`.
| `10a+3.5b=950` | ` xx 1->` | | `` | `10a` | `+` | `3.5b` | | | `=` | `950` | `` |
| | − | |
| `5a+b=400` | ` xx 3.5->` | | `` | `17.5a` | `+` | `3.5b` | | | `=` | `1400` | `` |
| | |
|
| | | `-` | `7.5a` | | | | | `=` | `-450` | ` -> (5)` |
Now use back substitution method
From (5)
`-7.5a=-450`
`=>a=(-450)/(-7.5)=60`
From (4)
`5a+b=400`
`=>5(60)+b=400`
`=>b+300=400`
`=>b=400-300=100`
From (1)
`a+b+c=200`
`=>(60)+(100)+c=200`
`=>c+160=200`
`=>c=200-160=40`
Solution using Elimination method.
`a = 60,b = 100,c = 40`
Thus, the missing frequencies are `60,100 and 40` respectively.
This material is intended as a summary. Use your textbook for detail explanation.
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