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Mathematics

The mean of the following frequency distribution is 62.8. Find the value of p :

ClassesFrequency
0 - 205
20 - 408
40 - 60p
60 - 8012
80 - 1007
100 - 1208

Measures of Central Tendency

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Answer

We construct the following table :

ClassesClass mark (ui)Frequency (fi)fiui
0 - 2010550
20 - 40308240
40 - 6050p50p
60 - 807012840
80 - 100907630
100 - 1201108880
Total40 + p2640 + 50p

Mean=ΣfiuiΣfi62.8=2640+50p40+p62.8(40+p)=2640+50p2512+62.8p=2640+50p62.8p50p=2640251212.8p=128p=10.\text{Mean} = \dfrac{Σfiui}{Σf_i} \\[1em] \Rightarrow 62.8 = \dfrac{2640 + 50p}{40 + p} \\[1em] \Rightarrow 62.8(40 + p) = 2640 + 50p \\[1em] \Rightarrow 2512 + 62.8p = 2640 + 50p \\[1em] \Rightarrow 62.8p - 50p = 2640 - 2512 \\[1em] \Rightarrow 12.8p = 128 \\[1em] \Rightarrow p = 10.

Hence, the value of p = 10.

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