Question

the median of the following data is 14.4 find the missing frequency x

Answer 1

Solution:

To find the value of x,first draw the table ,and calculate Cumulative frequency,as shown below

$\begin{tabular}{|c|c|c|}\cline{1-3}Class\: Interval&Freq&CF\\\cline{1-3}0-6&4&4\\\cline{1-3}6-12&x&\bf 4+x\\\cline{1-3}\bf 12-18&\bf 5&9+x\\\cline{1-3} 18-24&6&15+x\\\cline{1-3}24-30&1&16+x\\\cline{1-3}Total&16+x&\\\cline{1-3}\end{tabular}$

Median class:

To find Median class divide the last entry of CF by 2,

(16+x)/2 = 8+x/2

so just upper class of 8+x/2 is 12-18

So,12-18 is median class

Formula:

$\bf Median = l + ( \frac{ \frac{n}{2} - cf}{f} )h \\ \\ 14.4 = 12 + \bigg(\frac{ \frac{16 + x}{2} -4-x }{5}\bigg)6\\\\2.4= \bigg(\frac{ \frac{16 + x-8-2x}{2}}{5}\bigg)6\\\\2.4=\bigg(\frac{8-x}{5}\bigg)3\\\\0.8\times5=8-x\\\\4=8-x\\\\x=4$

So, x= 4

Mode: After placing the value of x,table becomes

$\begin{tabular}{|c|c|}\cline{1-2}Class\: Interval&Freq\\\cline{1-2}0-6&4\\\cline{1-2}6-12&4\\\cline{1-2}12-18&5\\\cline{1-2} \bf 18-24&6\\\cline{1-2}24-30&1\\\cline{1-2}Total&20&\\\cline{1-2}\end{tabular}$

Modal class :18-24

$\bf Mode= l+\bigg(\frac{f1-fo}{2f1-fo-f2}\bigg)h\\\\=18+\bigg(\frac{6-5}{12-5-1}\bigg)6\\\\=18+1\\\\Mode=19$

Hope it helps you