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Find the mean, mode and median of the following frequency distribution:
| Class: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |
| Frequency: | 4 | 4 | 7 | 10 | 12 | 8 | 5 |
Given: Here given a table of class and frequency.
To do: To find out mean, mode and median.
Solution:
| Class | $f_{i}$ | Class mark$(x_{i})$ | $f_{i}x_{i}$ | Cumulative frequency |
| 0-10 | 4 | 5 | 20 | 4 |
| 10-20 | 4 | 15 | 60 | 8 |
| 20-30 | 7 | 25 | 175 | 15 |
| 30-40 | 10 | 35 | 350 | 25 |
| 40-50 | 12 | 45 | 540 | 37 |
| 50-60 | 8 | 55 | 440 | 45 |
| 60-70 | 5 | 65 | 325 | 50 |
$\Sigma f_{i} =50$ | $\Sigma f_{i} x_{i} =1910$ |
Mean$=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}$
$=\frac{1910}{50}$
$=38.2$
Thus, the mean of the given data is 38.2.
Here $n=50$
$\frac{n}{2} =\frac{50}{2} =25$
Cumulative frequency just greater than 25 is 37 and the corresponding class is 40-50.
We take medium class 40-50.
Median $m=l+\left(\frac{\left(\frac{n}{2} -c.f.\right)}{f}\right)\times h$
M$=40+\left(\frac{25-15)}{32}\right)\times 10$ $( l=40,f=10, c.f.=15, h=10)$
$=40$
Thus, the median is 40.
We know that,
$Mode=3( median)-2( mean)$
$=3\times 40-2\times 38.2$
$=120-76.4$
$=43.6$
Hence, Mean$=38.2$, Median$=40$ and Mode$=43.6$.
Updated on: 10-Oct-2022
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