Question

Mean of the following data is 20 then, find p.
X
15
17
19
20+ P
23
f
2
3
4
5p
6
rks :​

Answer 1

$\large\underline{\sf{Solution-}}$

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c|c}\sf x&\sf \: f&\sf \:fx\\\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}&\frac{\qquad \qquad}{}\\\sf 15&\sf 2&\sf30\\\\\sf 17&\sf 3&\sf51\\\\\sf 19 &\sf 4&\sf76\\\\\sf 20 + p&\sf 5p&\sf100p + {5p}^{2} \\\\\sf 23&\sf 6&\sf138\\\frac{\qquad}{}&\frac{\qquad}{}&\frac{\qquad \qquad}{}\\\sf & \sf & \end{array}}\end{gathered}\end{gathered}\end{gathered}$

$\rm :\longmapsto\: \sum \: f \: = \: 15 + 5p$

$\rm :\longmapsto\: \sum \: fx \: = \: 295 + 100p + {5p}^{2}$

$\rm :\longmapsto\: \overline{x} \: = \: 20$

We know that,

Mean using direct method is given by

$\rm :\longmapsto\: \overline{x} \: = \: \dfrac{ \sum \: fx}{ \sum \: f}$

$\rm :\longmapsto\:20 = \dfrac{295 + 100p + {5p}^{2} }{15 + 5p}$

$\rm :\longmapsto\:300 + \cancel{100p} = 295 + \cancel{100p} + {5p}^{2}$

$\rm :\longmapsto\: {5p}^{2} = 5$

$\rm :\longmapsto\: {p}^{2} = 1$

$\bf\implies \:p = 1$

Additional Information :-

Mean using Short Cut method is given by

$\rm :\longmapsto\: \overline{x} \: = A + \: \dfrac{ \sum \: fd}{ \sum \: f}$

Mean using Step Deviation method is given by

$\rm :\longmapsto\: \overline{x} \: = A + \: \dfrac{ \sum \: fu}{ \sum \: f} \times h$