mean is 145.20 If true then enter 1 and if false then enter 0
A
non of thease
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B
0
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C
1
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D
can not det
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Solution
The correct option is C 1 ForgroupeddatawhenClassIntervalisgiven,wefindtheClassMark.ClassMark=midpointofaninterval=(Lowerlimit+Upperlimit)2ClassMarkistakenasxiforeachclassinterval.Findxifi.Fromthetable,wehaveΣxifi=7260&Σfi=50Mean=ΣxifiΣfi=726050=145.20