Question

plzzz solve this question​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

$<!-- Codes by HTML.am --></p><p></p><p><!-- CSS Code --></p><p><style type="text/css" scoped></p><p>table.GeneratedTable {</p><p>width:100%;</p><p>background-color:#FFFFFF;</p><p>border-collapse:collapse;border-width:1px;</p><p>border-color:#336600;</p><p>border-style:solid;</p><p>color:#009900;</p><p>}</p><p></p><p>table.GeneratedTable td, table.GeneratedTable th {</p><p>border-width:1px;</p><p>border-color:#336600;</p><p>border-style:solid;</p><p>padding:3px;</p><p>}</p><p></p><p>table.GeneratedTable thead {</p><p>background-color:#CCFF99;</p><p>}</p><p></style></p><p></p><p><!-- HTML Code --></p><p><table class="GeneratedTable"></p><p><thead></p><p><tr></p><p><th>Class Interval</th></p><p><th>Frequency</th></p><p><th>Step-arithmatic frequency (Lower index)</th></p><p></tr></p><p></thead></p><p><tbody></p><p><tr></p><p><td>0-100</td></p><p><td>2</td></p><p><td>2</td></p><p></tr></p><p><tr></p><p><td>100-200</td></p><p><td>5</td></p><p><td>2+5=7</td></p><p></tr></p><p><tr></p><p><td>200-300</td></p><p><td>x</td></p><p><td>7+x</td></p><p></tr></p><p><tr></p><p><td>300-400</td></p><p><td>12</td></p><p><td>7+x+12=19+x</td></p><p></tr></p><p><tr></p><p><td>400-500</td></p><p><td>17</td></p><p><td>19+x+17=36+x</td></p><p></tr></p><p><tr></p><p><td>500-600</td></p><p><td>20</td></p><p><td>36+x+20=56+x</td></p><p></tr></p><p><tr></p><p><td>600-700</td></p><p><td>y</td></p><p><td>56+x+y</td></p><p></tr></p><p><tr></p><p><td>700-800</td></p><p><td>9</td></p><p><td>56+x+y+9=65+x+y</td></p><p></tr></p><p><tr></p><p><td>800-900</td></p><p><td>7</td></p><p><td>65+x+y+7=72+x+y</td></p><p></tr></p><p><tr></p><p><td>900-1000</td></p><p><td>4</td></p><p><td>72+x+y+4=76+x+y</td></p><p></tr></p><p><tr></p><p><td> </td></p><p><td>N=100</td></p><p><td> </td></p><p></tr></p><p></tbody></p><p></table>$

Now given that .....

Total frequency=N=100

$\therefore$$\implies 76+x+y=100\\</p><p>\implies x+y=100-76\\ \implies\boxed{ x+y=24}.........(i)$

Now the number of terms is =N=100

$\therefore$$Middle\: class \:term \:is =\frac{N}{2}=\frac{100}{2}=50&#39;th\: term$

$\therefore \bf median &#45;</p><p>class\: is =500-600$

$\therefore \boxed{ \bf\red{median=l + (\frac{ \frac{n}{2} -c_f }{f} ) \times h}}$

$\bf\rightarrow l=lower\: class\: boundary\: of\: median \:class\:\\ \bf\rightarrow n=number\: of\: frequency\\ \bf\rightarrow f=number\: of\: frequency\: of\: median \:class \\ \bf\rightarrow h=class \:length\: of\: median\: class\\ \bf \rightarrow c_f=step\: arithmetic \:frequency\: of\: the\:\\ \bf previous\: class\: just \:the \:median\: class.$

•Given(from the data frequency table):-

$\rightarrow l=500\\ \rightarrow n=100\\</p><p>\rightarrow f=20\\ \rightarrow h=100\\</p><p>\rightarrow c_f=36+x\\ \rightarrow median=525(given)$

•Putting the values:-

$\therefore \bf\red{median=500+ (\frac{ \frac{100}{2} -(36+x)}{20} ) \times 100}\\ \implies \bf\red{525=500+ (\frac{ 50 -36-x)}{\cancel{20} }) \times \cancel{100}}\\ </p><p> \implies \bf\red{525-500= (14-x)\times 5}\\ \implies 25=70-5x\\ \implies 5x=70-25\\ \implies x=\frac{45}{5}\\ \implies \boxed{\red{x=9} }$

Now from equation (i)

$\implies x+y=100\\ \implies 9+y=24\\ \implies y=24-9\\ \implies\boxed{\red{ y=15}}$

$\therefore \bf x=9$

$\therefore \bf y=15$

______________________________________________

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

$\mathcal{ \&#35;\mathcal{answer with quality }\: \: \&#38; \: \: \&#35;BAL }$