Find the median for the following frequency distribution.
Class Intervals
$160 - 162$ $163 - 165$ $166 - 168$ $169 - 171$ $172 - 174$
frequency $15$ $117$ $136$ $110$ $14$

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Solution

$\Rightarrow$ $\frac{N}{2} = \frac{392}{2} = 196$

Just bigger commulative frequency $268$ and interval $(165.5 - 168.5)$
$\Rightarrow$ $M e d i a n (M) = l + \frac{(\frac{N}{2} - F)}{f} \times x_{i} = 165.5 + \frac{(196 - 132)}{136} \times 3 = 166.9$

Hence, this is the answer.

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Q. Find the median for the following frequency distribution

Height (in $c m$ )	$160 - 162$	$163 - 165$	$166 - 168$	$169 - 171$	$172 - 174$
Frequency	$15$	$117$	$136$	$110$	$14$

Now look at this problem: the heights of 40 people, in centimeters, are given below. Make a frequency table of this data.

175	143	177	163	170	168	156	168	145	160
159	154	148	166	168	150	164	160	176	163
140	170	165	168	169	174	158	148	155	165
162	167	171	165	171	167	159	171	167	180

The following is the distribution of height of students of a certain class in a certain city:
Height (in cms): 160-162 163-165 166-168 169-171 172-174
No. of students: 15 118 142 127 18
Find the average height of maximum number of students.

Question 4 (i)
The heights of 50 students, measured to the nearest centimeters, have been found to be as follows:
$\begin{matrix} 161 & 150 & 154 & 165 & 168 & 161 & 154 & 162 & 150 & 151 162 & 164 & 171 & 165 & 158 & 154 & 156 & 172 & 160 & 170 153 & 159 & 161 & 170 & 162 & 165 & 166 & 168 & 165 & 164 154 & 152 & 153 & 156 & 158 & 162 & 160 & 161 & 173 & 166 161 & 159 & 162 & 167 & 168 & 159 & 158 & 153 & 154 & 159 \end{matrix}$
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 - 165, 165 - 170, etc.

Question 4 (i)
The heights of 50 students, measured to the nearest centimeters, have been found to be as follows:
$\begin{matrix} 161 & 150 & 154 & 165 & 168 & 161 & 154 & 162 & 150 & 151 162 & 164 & 171 & 165 & 158 & 154 & 156 & 172 & 160 & 170 153 & 159 & 161 & 170 & 162 & 165 & 166 & 168 & 165 & 164 154 & 152 & 153 & 156 & 158 & 162 & 160 & 161 & 173 & 166 161 & 159 & 162 & 167 & 168 & 159 & 158 & 153 & 154 & 159 \end{matrix}$
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 - 165, 165 - 170, etc.

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Class Intervals	$160 - 162$	$163 - 165$	$166 - 168$	$169 - 171$	$172 - 174$
frequency	$15$	$117$	$136$	$110$	$14$

175	143	177	163	170	168	156	168	145	160
159	154	148	166	168	150	164	160	176	163
140	170	165	168	169	174	158	148	155	165
162	167	171	165	171	167	159	171	167	180

175	143	177	163	170	168	156	168	145	160
159	154	148	166	168	150	164	160	176	163
140	170	165	168	169	174	158	148	155	165
162	167	171	165	171	167	159	171	167	180

Find the median for the following frequency distribution.Class Intervals 160−162163−165166−168169−171172−174frequency 1511713611014

⇒N2=3922=196 Just bigger commulative frequency 268 and interval (165.5−168.5)⇒ Median(M)=l+(N2−F)f×xi=165.5+(196−132)136×3=166.9Hence, this is the answer.

Find the median for the following frequency distribution.
Class Intervals
$160 - 162$ $163 - 165$ $166 - 168$ $169 - 171$ $172 - 174$
frequency $15$ $117$ $136$ $110$ $14$

$\Rightarrow$ $\frac{N}{2} = \frac{392}{2} = 196$

Just bigger commulative frequency $268$ and interval $(165.5 - 168.5)$
$\Rightarrow$ $M e d i a n (M) = l + \frac{(\frac{N}{2} - F)}{f} \times x_{i} = 165.5 + \frac{(196 - 132)}{136} \times 3 = 166.9$

Hence, this is the answer.

175	143	177	163	170	168	156	168	145	160
159	154	148	166	168	150	164	160	176	163
140	170	165	168	169	174	158	148	155	165
162	167	171	165	171	167	159	171	167	180