Question

Find the median:
Class Interval Frequency
160-162 15
163-165 118
166-168 142
169-171 127
172-174 18

Answer 1

Answer :

We form our data table :

Class interval	Mid point ( x )	Frequency ( f )	fx	cumulative frequency cfb
160 - 162	161	15	2415	15
163 - 165	164	118	19352	15 + 118 = 133
166 - 168	165	142	23430	133 + 142 = 275
169 - 171	170	127	21590	275 + 127 = 402
172 - 174	173	18	3114	402 + 18 = 420
		${\sum }_{}$f = 420	${\sum }_{}$fx = 69901

So,

Mean = $\frac{{\sum }_{}fx}{{\sum }_{}f}=\frac{69901}{420}=\mathbf{166}\mathbf{.}\mathbf{43}$
So,
Median group = 166 - 168
And
We know Median = $L+\frac{{\displaystyle \frac{n}{2}}-c{f}_b}{f_m}\times w$
Here

L = The lower class boundary of the group containing the median = 166
n = The total number of data = 420
cf_b = The cumulative frequency of the groups before the median group = 133
f_m = The frequency of the median group = 142
w = The group width = 3

So,

Median = $166+\frac{{\displaystyle \frac{420}{2}-133}}{142}\times 3=166+\frac{{\displaystyle 210-133}}{142}\times 3=166+\frac{{\displaystyle 77\times 3}}{142}=166+\frac{{\displaystyle 231}}{142}=166+1.6267=167.6267\approx \mathbf{167}\mathbf{.}\mathbf{63}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{}\mathit{A}\mathit{n}\mathit{s}\mathbf{}\mathbf{)}\mathbf{}$