First, we construct a cumulative frequency table
`therefore (N)/(2)=(280)/(2)=140`
(i) Here, median class is 10-15, because 140 lies in it.
Lower limit (l)=10, Frequency (f)=133.
Cumulative frequency(cf)=49 and class width (h)=5
`therefore "Median"=l+(((N)/(2)-cf))/(f)xxh`
`=10+((140-49))/(133)xx5`
`=10+(91xx5)/(133)`
`=10+(455)/(133)=10+3.421`
`=13.421xx1000`
=13421
(ii) Here, the highest frequency is 133, which lies 10-15, called class Lower, limit (l)=10 class width (h)=`5,f_(m)=133, f_(1)=49 and f_(2)=63`
`therefore "Mode"=l+((f_(m)+f_(1))/(2f_(m)-f_(1)-f_(2)))xxh`
`=10+{(133-49)/(2xx133-49-63)}xx5`
`=10+(84xx5)/(266-112)=10+(84xx5)/(154)=10+2,727`
=12,727(in thousand)
=`12727xx1000=12727`
Hence, the meidan and modal salary are 13421 adn 12727 respectively,
