Question

The median of the distribution given below is 14.4. Find the values of x and y. If the total frequency is 20​

Answer 1

$\huge\bf\red{\underline{\underline{Given}}}\::$

$\begin{cases}\sf\gray{The \ median \ of \ the \ distribution \ is \ 14.4} \\ \sf\gray{Total \ frequency \ is \ 20}\end{cases}$

$\huge\bf\green{\underline{\underline{To\:Find}}}\::$

$\begin{cases}\sf\gray{Find \ the \ value \ of \ x \ and \ y}\end{cases}$

$\huge\bf\purple{\underline{\underline{Solution}}}\::$

$<TABLE BORDER="15"    WIDTH="10%"   CELLPADDING="15" CELLSPACING="1"></p><p>   </p><p>   <TR></p><p>      <TH>ᏟᏞᎪՏՏ ᏆΝͲᎬᎡᏙᎬᏞ</TH></p><p>   <TH>ҒᎡᎬϘႮᎬΝᏟᎽ</TH></p><p>  <TH>ᑕ.ᖴ.</TH></p><p></p><p>   <TR ALIGN="CENTER"></p><p>      <TD>0-6</TD></p><p>      <TD>4</TD> </p><p><TD>4</TD></p><p></p><p></TR><TR ALIGN="CENTER"></p><p>      <TD>6-12</TD></p><p>      <TD>x</TD> <TD>4+x</TD> </p><p>   </TR> </p><p></p><p>  <TR ALIGN="CENTER"></p><p>      <TD>12-18</TD></p><p>      <TD>5</TD> </p><p> <TD>9+x</TD></p><p>   </TR> </p><p></p><p><TR ALIGN="CENTER"></p><p>      <TD>18-24</TD></p><p>      <TD>y</TD> </p><p><TD>9+x+y</TD></p><p></TR> </p><p></p><p><TR ALIGN="CENTER"></p><p>      <TD>24-30</TD></p><p>      <TD>1</TD> </p><p><TD>10+x+y</TD></p><p>   </TR> </p><p></p><p> <TR ALIGN="CENTER"></p><p>      <TD></TD></p><p>      <TD>∑⨏ = 20</TD> </p><p><TD></TD></p><p>    </TR> </p><p></p><p></TABLE>$

✶ $\sf\orange{10 \ + \ x \ + \ y \ = \ 20}$

$→ \:\: \sf\blue{x \ + \ y \ = \ 20 \ - \ 10}$

$→ \:\: \sf\orange{x \ + \ y \ = \ 10 \ ........(1)}$

$\bf\purple{Hence, \ Median \ class \ is \ 12-18}$

$\begin{cases}\sf\gray{Lower \ limit \ of \ the \ Median \ Class,\: l \: = \ 12} \\ \sf\gray{Class \ width,\: h \ = \ 6} \\ \sf\gray{C.F. \ = \ 4+x} \\ \sf\gray{Frequency \ of \ the \ median \ class \ = \ 5}\\ \sf\gray{\frac{n}{2} \ = \ \frac{20}{2} \ = \ 10}\end{cases}$

★ ${\boxed{\bf{\red{Median \ = \ l \ + \ \dfrac{\frac{n}{2} \ - \ cf}{f} \ \times \ h}}}}$

$\longrightarrow\:\:\sf\purple{14.4 \ = \ 12 \ + \ \dfrac{10 \ - \ (x \ + \ 4)}{5} \ \times \ 6}$

$\longrightarrow\:\:\sf\green{14.4 \ = \ 12 \ + \ \dfrac{6 \ \times \ (10 \ - \ x \ - \ 4}{5}}$

$\longrightarrow\:\:\sf\purple{12 \ = \ 36 \ - \ 6x}$

$\longrightarrow\:\:\sf\green{6x \ = \ 36 \ - \ 12}$

$\longrightarrow\:\:\sf\purple{6x \ = \ 24}$

$\longrightarrow\:\:\sf\green{x \ = \ \dfrac{\cancel{24}}{\cancel{6}}}$

$\longrightarrow\:\:\sf\underline\red{x \ = \ 4}$

$\dag\:\:\bf\underline\pink{Putting \ x \ value \ in \ eq \ (1)}$

$\mapsto\:\:\sf\purple{x \ + \ y \ = \ 10}$

$\mapsto\:\:\sf\green{ 4 \ + \ y \ = \ 10}$

$\mapsto\:\:\sf\purple{y \ = \ 10 \ - \ 4}$

$\mapsto\:\:\sf\underline\red{y \ = \ 6}$

$\bf\orange{Hence}\begin{cases}\bf\underline\red{\:x \ = \ 4 \:}\\ \bf\underline\red{\:y \ = \ 6\:}\end{cases}$