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Question

Mean life of following distribution in the following distribution is 57.6 hours. If sum of frequency is 50 then find the missing frequency x and y

0207
2040x
406012
6080y
801008
100120 5



A
x=7,y=11
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B
x=11,y=7
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C
x=13,y=5
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D
x=8,y=10
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Solution

The correct option is B x=8,y=10
C.I Frequency
fi
Midvalue
xi
fixi
020 7 10 70
2040 x 30 30x
4060 12 50 600
6080 y 70 70y
80100 8 90 720
100120 5 110 550
fi=32+x+y fixi=1940+30x+70y
fi=32+x+y
50=32+x+y
x+y=18 ------- ( 1 )
¯¯¯x=fixifi

57.6=1940+30x+70y50

2880=1940+30x+70y
30x+70y=940
3x+7y=94 ----- ( 2 )
Multiplying equation ( 1 ) by 3 we get,
3x+3y=54 ------ ( 3 )
Subtracting equaton ( 3 ) from equation ( 2 ) we get,
4y=40
y=10
Substituting y=10 in equation ( 1 ),
x+10=18
x=8
x=8 and y=10

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