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Mathematics

If the mean of following distribution is 681368\dfrac{1}{3}. Find the missing frequency p.

C.I.f
25-3510
35-456
45-554
55-65p
65-754
75-8512
85-9526

Statistics

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Answer

C.I.fClass mark (x)fx
25-351030300
35-45640240
45-55450200
55-65p6060p
65-75470280
75-851280960
85-9526902340
TotalΣf = 62 + pΣfx = 4320 + 60p

By formula,

Mean = ΣfxΣf\dfrac{Σfx}{Σf}

Given,

Mean = 681368\dfrac{1}{3}

6813=4320+60p62+p2053=4320+60p62+p205(62+p)=3(4320+60p)12710+205p=12960+180p205p180p=129601271025p=250p=25025=10.\therefore 68\dfrac{1}{3} = \dfrac{4320 + 60p}{62 + p} \\[1em] \Rightarrow \dfrac{205}{3} = \dfrac{4320 + 60p}{62 + p} \\[1em] \Rightarrow 205(62 + p) = 3(4320 + 60p) \\[1em] \Rightarrow 12710 + 205p = 12960 + 180p \\[1em] \Rightarrow 205p - 180p = 12960 - 12710 \\[1em] \Rightarrow 25p = 250 \\[1em] \Rightarrow p = \dfrac{250}{25} = 10.

Hence, p = 10.

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