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The Following Process Might Help,



Classes ......................... Mid ....................... Frequency ........... Deviation........ fd(F Multiply D)

2 h 50 m  t  3 h 00 m .... 2 h 55 m ....................5 ........................-10 m...................... -50 m
3 h 00 m  t  3 h 10 m .... 3 h 5 m (a) .................14 .........................0 m
3 h 10 m  t  3 h 20 m .... 3 h 15 m .....................7 ..........................10 m......................... 70 m
3 h 20 m  t  3 h 30 m .... 3 h 25 m .....................4.......................... 20 m....................... 80 m



So Sum up fd which = 100
then, Divide by total no of Frequency which is = 30



So on . Results would be 3. 33333333333333333333333333333
Just round to nearest min= 3 m
and then add to the the assumed mean



So you have a answer, 3 h 8 m

Tell me if helps,
Kind Regards ,
Asad


thank alot dude
 
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