Here, we observe that, all 100 residents of a town have age equal and above 0, Since 90 resisdents of a town have age equal and above 10. So,100-90=10 resisdents lies in the interval 0-10 and so on. Continue in this manner, we get frequency of all class intervals. Now we construct the frequency distribution table.
Here, (assumed mean)a=35
and class width h=10
By step deviation method.
`"Mean"(bar(x))=a+(sumf_(i)u_(i))/(sumf_(i))xxh`
`35+((-40))/(100)xx10`
`=35-4=31`
Hence, the required mean age is 31 yr.
