Question

If the mean of the following data is 18.75, find the value of p

Xi = 10,15,p,25,30 and Fi = 5,10,7,8,2

Answer 1

AnswEr :

$\begin{tabular}{| c | c | c | c | c | c | }\cline{1-6}</p><p>\bf{xi} & \sf{10} & \sf{15} & \sf{p} & \sf{25} & \sf{30}\\ \cline{1-6}\bf{fi} & \sf{5} & \sf{10} & \sf{7} & \sf{8} & \sf{2}\\ \cline{1-6} \end{tabular}$

$\rule{100}1$

$\begin{tabular}{| c | c | c | }\cline{1-3}</p><p>\bf{xi} & \bf{fi} & \bf{fixi}\\ \cline{1-3}\sf{10} & \sf{5} & \sf{50}\\ \cline{1-3}\sf{15} & \sf{10} & \sf{150}\\ \cline{1-3}\sf{p} & \sf{7} & \sf{7p}\\ \cline{1-3}\sf{24} & \sf{8} & \sf{200}\\ \cline{1-3}\sf{30} & \sf{2} & \sf{60}\\ \cline{1-3}\sf{Total } & \sum\limits\sf\ fi \sf\ = 30 & \sum\limits\sf\ fixi \sf\ = 460 + 7p \\ \cline{1-3} \end{tabular}$

$\rule{200}2$

$\normalsize\maltese \: \: {\boxed{\sf{Mean \left( \bar{x} \right) = \frac{\sum\limits\ f_i x_i}{\sum\limits\ f_i} }}}$

$\normalsize\dashrightarrow\sf\ 18.75 = \frac{460 + 7p}{32}$

$\scriptsize\sf{ \: \: \: \: \: \: \: \: \: \dag\ Cross \: multiplication}$

$\normalsize\dashrightarrow\sf\ 18.75(32) = 460 + 7p$

$\normalsize\dashrightarrow\sf\ 600 = 460 + 7p$

$\normalsize\dashrightarrow\sf\ 600 - 460 = 7p$

$\normalsize\dashrightarrow\sf\ 140 = 7p$

$\normalsize\dashrightarrow\sf\ p = \frac{\cancel{140}}{\cancel{7}}$

$\normalsize\dashrightarrow\sf\ p = 20$

$\therefore\:\underline{\textsf{Hence, \: the \: value \: of \: p \: is}{\textbf{\: \: 20 }}}$

Answer 2

${\underline{\underline{\mathbb{\orange{ REFER \: TO\: THE \: ATTACHMENTS! }}}}}$

☺️

$\Huge\mathfrak{Or}$

$\sf\pink{↪\sum fi = 32}$

$\sf\red{↪\sum fixi = 460+7p}$

$\sf\blue{ ↪ Mean = 18.75 }$

$\sf\orange{↪Mean = \frac{ \sum{fixi} }{\sum{fi} } }$

$\sf\purple{↪18.75= \frac{460+7p}{32} }$

$\sf{On \: cross \: multiplication :}$

$\sf\green{↪18.75 \times 32 = 460 + 7p}$

$\sf{600-460 = 7p}$

$\sf\purple{ ↪ \cancel\frac{140}{7} = p}$

$\boxed{\sf{\orange{↪p = 20}}}$