Find missing frequency from the following data
| Class | Frequency |
| 0 - 400 | 14 |
| 400 - 800 | 22 |
| 800 - 1200 | ? |
| 1200 - 1600 | 124 |
| 1600 - 2000 | ? |
| 2000 - 2400 | 32 |
| 2400 - 2800 | 15 |
| 2800 - 3200 | 5 |
Total Frequency (N) = 360 and mode = 1376
Solution:
Class
`(1)` | Frequency `(f)`
`(2)` |
| 0 - 400 | 14 |
| 400 - 800 | 22 |
| 800 - 1200 | a`=f_0` |
| 1200 - 1600 | 124`=f_1` |
| 1600 - 2000 | b`=f_2` |
| 2000 - 2400 | 32 |
| 2400 - 2800 | 15 |
| 2800 - 3200 | 5 |
| --- | --- |
| `n=212+a+b` |
`n = 360`
`212 + a+b= 360`
`a+b=148 ->(1)`
To find Mode Class
Here, mode is `1376`.
`:.` The mode class is `1200 - 1600`.
`:. L = `lower boundary point of mode class `=1200`
`:. f_1 = ` frequency of the mode class `=124`
`:. f_0 = ` frequency of the preceding class `=a`
`:. f_2 = ` frequency of the succedding class `=b`
`:. c = ` class length of mode class `=400`
`Z = L + ((f_1 - f_0) / (2*f_1 - f_0 - f_2)) * c`
`1376=1200 + ((124 - a)/(2*124 - a - b)) * 400`
`1376 - 1200 = ((124 - a)/(2*124 - a - b)) * 400`
`176=((124 - a)/(2*124 - a - b)) * 400`
`176=((124 - a)/(-a-b+248)) * 400`
`176*(-a-b+248)=(124-a)*400`
`-176a-176b+43648=-400a+49600`
`224a-176b=5952`
`16(14a-11b)=16*372`
`14a-11b=372 ->(2)`
Now solving this 2 equations using substitution method
`a+b=148`
and `14a-11b=372`
Suppose,
`a+b=148 ->(1)`
and `14a-11b=372 ->(2)`
Taking equation `(1)`, we have
`a+b=148`
`=>a=-b+148 ->(3)`
Putting `a=-b+148` in equation `(2)`, we get
`14a-11b=372`
`14(-b+148)-11b=372`
`=>-14b+2072-11b=372`
`=>-25b+2072=372`
`=>-25b=372-2072`
`=>-25b=-1700`
`=>b=68 ->(4)`
Now, Putting `b=68` in equation `(3)`, we get
`a=-b+148`
`=>a=-1(68)+148`
`=>a=-68+148`
`=>a=80`
`:.a=80" and "b=68`
Thus, the missing frequencies are `80 and 68` respectively.
This material is intended as a summary. Use your textbook for detail explanation.
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