Question

Find the value of x and y.

Thanks ❤️ :D​

Answer 1

GivEn:

$\normalsize\;\;\bullet\;\;\sf Median\;of\;the\;given\;data\;:\; \bf{525}\\\\ \normalsize\;\;\bullet\;\;\sf Total\; Frequency\;:\; \bf{100}$

To find:

$\normalsize\;\;\bullet\;\;\sf Values\;of\;x\;and\;y$

Solution:

$\boxed{\begin{array}{cccc}\sf Class\: interval&\sf Frequency&\sf C.F\\\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad}{}&\frac{\qquad \qquad \qquad \qquad\qquad}{}\\\sf 0-100&\sf 2&\sf 2 \\\\\sf 100-200 &\sf 5&\sf 7 \\\\\sf 200-300 &\sf x &\sf 7 + x \\\\\sf 300-400&\sf 12&\sf 19 + x\\\\\sf 400-500 &\sf 17 &\sf 36 + x \\\\\sf 500-600 &\sf 20 &\sf 56 + x \\\\\sf 600-700 &\sf y &\sf 56 + x + y \\\\\sf 700-800 &\sf 9 &\sf 65 + x + y \\\\\sf 800-900 &\sf 7 &\sf 72 + x + y \\\\\sf 900-1000 &\sf 4 &\sf 76 + x + y\end{array}}$

Here,

$\maltese\;{\boxed{\sf{Median = 525}}}$

And, 500 - 600 is median class

Where,

$\normalsize\;\;\bullet\;\;\sf Lower\;Limit\;of\;median\;class\;(l)\;:\; \bf{500}\\\\ \normalsize\;\;\bullet\;\;\sf Class\;interval\;(h)\;:\; 100 - 0\;:\; \bf{100}\\\\ \normalsize\;\;\bullet\;\;\sf n\;:\; \bf{100}\\\\ \normalsize\;\;\bullet\;\;\sf Cumulative\; frequency\;(CF)\;:\; \bf{36 + x}$

$\rule{170}{2}$

$\underline{\bigstar\:\textbf{According to the Question :}}\\\\ \sf As\;we\;know\;that,\\\\ \maltese\;{\boxed{\sf{Median = l + \dfrac{ \frac{n}{2} - cf}{f} \times h}}}\\\\ {\underline{\sf{\bigstar\;Put\;the\;given\;values\;:}}}\\\\ :\implies\sf 525 = 500 + \dfrac{ \frac{100}{2} - (36 - x)}{ \cancel{20}} \times \cancel{100}\\\\ :\implies\sf 525 = 500 + [50 - (36 - x)] \times 5$

$:\implies\sf 525 - 500 = (50 - 36 - x) \times 5\\\\ :\implies\sf 25 = (14 - x) \times 5\\\\ :\implies\sf 25 = 70 - 5x\\\\ :\implies\sf 5x = 70 - 25\\\\ :\implies\sf 5x = 45\\\\ :\implies\sf x = \cancel{ \dfrac{45}{5}}\\\\ :\implies{\underline{\boxed{\sf{\purple{x = 9}}}}}\;\bigstar$

$\sf Also\;we\;have,\\\\ :\implies\sf \sum fi = 76 + x + y\\\\ :\implies\sf 100 = 76 + x + y\\\\ :\implies\sf 100 - 76 = x + y\\\\ :\implies\sf 24 = 9 + y\\\\ :\implies\sf y = 24 - 9\\\\ :\implies{\underline{\boxed{\sf{\pink{y = 15}}}}}\;\bigstar\\\\ \therefore\;\sf \underline{The\;value\;of\;x\;and\;y\;are\;9\;and\;15\; respectively}$

Answer 2

Solution :

Data given :

$\begin{tabular}{|c|c|c|} \cline{1-3}\multicolumn{3}{|c|} {DATA} \\ \cline{1-3} \bf Class-Interval & \bf Frequency(f_i) &\bf Cumulative-frequency \\ \cline{1-3} 0-100 & 2 & 2\\ \cline{1-3} 100-200& 5&7\\ \cline{1-3} 200-300 & x & 7+x \\ \cline{1-3} 300-400 & 12 & 19+x \\ \cline{1-3} 400-500 & 17&36+x\\ \cline{1-3} 500-600& 20 &56+x\\\cline{1-3}600-700&y&56+x+y\\\cline{1-3}700-800&9&65+x+y\\\cline{1-3} 800-900&7&72+x+y\\\cline{1-3}900-1000 & 4&76+x+y\\\cline{1-3}\bf Total & 100 &\\ \cline{1-3} \end{tabular}}$

$\underline{\bf{Explanation\::}}}}$

$\bigstar$Firstly, we know that formula of the Median :

$\boxed{\bf{Median =l+\Bigg( \frac{\dfrac{n}{2} -cf}{f} \Bigg)\times h}}}$

Where as;

• n = number of observations.
• l = lower limit of median class.
• f = frequency of median class.
• cf = cumulative frequency.
• h = class size (height) .

$\underline{\boldsymbol{According\:to\:the\:question\::}}}$

Median = 525, l = 500, n = 100, cf = 36 + x , f = 20 & h = 100

$\longrightarrow\sf{76+x+y= 100}\\\\\longrightarrow\sf{x+y=100-76}\\\\\longrightarrow\sf{x+y=24..................(1)}$

&

$\longrightarrow\sf{525 = 500 +\Bigg[\dfrac{\cancel{\dfrac{100}{2}} -(36 + x)}{20}\Bigg] \times 100}\\\\\\\longrightarrow\sf{525 = 500 + \Bigg[\dfrac{50 - (36+ x)}{20} \Bigg]\times 100}\\\\\\\longrightarrow\sf{525 - 500 = \dfrac{50 - 36 - x}{\cancel{20}} \times \cancel{100}}\\\\\\\longrightarrow\sf{25=(14 - x )5}\\\\\longrightarrow\sf{25 =70 - 5x}\\\\\longrightarrow\sf{-5x=25-70}\\\\\longrightarrow\sf{-5x = -45}\\\\\longrightarrow\sf{x=\cancel{-45/-5}}\\\\\longrightarrow\bf{x=9}$

∴ Putting the value of x in equation (1),we get;

$\longrightarrow\sf{9 + y =24}\\\\\longrightarrow\sf{y=24-9}\\\\\longrightarrow\bf{y=15}$

Thus;

The value of x & y will be 9 & 15 .