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Find the value of P, if the mean of the following distribution is 18.

x:

13

15

17

19

20 + p

23

f:

8

2

3

4

5p


xi

fi

fixi

13

8

104

15

2

30

17

3

51

19

4

76

20 + p

5p

5p (20 + p)

23

6

138

 

Σfi = 23 + 5p

Σfixi = 399 + 5p2 + 100p

 

Here, we have                    top enclose straight x equals 18 comma space space sum from blank to blank of straight f space equals space 23 space plus space 5 straight p space and space sum from blank to blank of straight f subscript straight i straight x subscript straight i space equals 399 plus 5 straight p squared plus 100 straight p

Now,                                 top enclose straight x space equals space fraction numerator begin display style sum from blank to blank of end style f subscript i x subscript i over denominator begin display style sum from blank to blank of end style f subscript i end fraction equals fraction numerator 5 p squared plus 100 p plus 399 over denominator 5 p plus 23 end fraction

rightwards double arrow                                     18 equals fraction numerator 5 straight p squared plus 10 straight p plus 399 over denominator 5 straight p plus 23 end fraction
rightwards double arrow                      18(5p + 23) = 5p2 + 100p + 399
rightwards double arrow                      90p + 414 = 5p2 + 100p + 399
rightwards double arrow                      5p2 + 10p - 15 = 0
rightwards double arrow                      p2 + 2p - 3 = 0
rightwards double arrow                      p2 + 3p - p - 3 = 0
rightwards double arrow                     p(p + 3) - 1 (p + 3) = 0
rightwards double arrow                     (p - 1) (p + 3) = 0
rightwards double arrow                      p - 1 = 0 or p + 3 = 0
Since                   p =  1 or  p = - 3 neglected
Therefore,            p = 1 


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