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Mathematics

If the mean of the following distribution is 27, then the value of p is :

ClassFrequency
0 – 108
10 – 20p
20 – 3012
30 – 4013
40 – 5010
  1. 6

  2. 7

  3. 9

  4. 11

Measures of Central Tendency

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Answer

ClassFrequency (f)Class mark(x)fx
0 – 108540
10 – 20p1515p
20 – 301225300
30 – 401335455
40 – 501045450
Total∑ f = 43 + p∑ fx = 1245 + 15p

By formula,

Mean=fxf27=1245+15p43+p27(43+p)=1245+15p1161+27p=1245+15p27p15p=1245116112p=84p=8412p=7.\Rightarrow \text{Mean} = \dfrac{\sum fx}{\sum f} \\[1em] \Rightarrow 27 = \dfrac{1245 + 15p}{43 + p} \\[1em] \Rightarrow 27(43 + p) = 1245 + 15p \\[1em] 1161 + 27p = 1245 + 15p \\[1em] \Rightarrow 27p - 15p = 1245 - 1161 \\[1em] 12p = 84 \\[1em] \Rightarrow p = \dfrac{84}{12} \\[1em] \Rightarrow p = 7.

Hence, option 2 is the correct option.

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