First we construct a cumulative frequency table.
It is given that, n=600
`therefore (n)/(2)=(600)/(2)=300`
Since, cumulative frequency 440 lies in the interval 1000-2000.
Here, (lower meidan class) l=1000, f=190, cf=250(class width) j=1000
and total observation n=600
`"Median"=l({(n)/(2)-cf})/(f)xxh`
`=1000+(300-250)/(190)xx1000`
`=1000+(50)/(190)xx1000`
`=1000+(5000)/(19)`
=1000+263.15=1263.15
Hence, the median income is 1263.15.