Find the mode of the following frequency table:
Class Interval Frequency
$140 - 150$ $4$
$150 - 160$ $6$
$160 - 170$ $10$
$170 - 180$ $12$
$180 - 190$ $9$
$190 - 200$ $3$

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Solution

Since the class, $170 - 180$ has the maximum frequency.
Therefore class $170 - 180$ is the modal class.
Here $l_{1} = 170, h = 180 - 170 = 10, f_{1} = 12, f_{0} = 10, f_{2} = 9$
$∵$ Mode $Z = l_{1} + [\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}] \times h$
$\Rightarrow Z = 170 + [\frac{12 - 10}{2 \times 12 - 10 - 9}] \times 10$
$= 170 + \frac{2 \times 10}{24 - 9}$
$= 170 + 4$
$= 174$

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Class Interval	Frequency
$140 - 150$	$4$
$150 - 160$	$6$
$160 - 170$	$10$
$170 - 180$	$12$
$180 - 190$	$9$
$190 - 200$	$3$

Find the mode of the following frequency table:Class IntervalFrequency140−1504150−1606160−17010170−18012180−1909190−200 3

Since the class, 170−180 has the maximum frequency.Therefore class 170−180 is the modal class.Here l1=170,h=180−170=10,f1=12,f0=10,f2=9∵ Mode Z=l1+[f1−f02f1−f0−f2]×h⇒Z=170+[12−102×12−10−9]×10 =170+2×1024−9=170+4=174

Find the mode of the following frequency table:
Class Interval Frequency
$140 - 150$ $4$
$150 - 160$ $6$
$160 - 170$ $10$
$170 - 180$ $12$
$180 - 190$ $9$
$190 - 200$ $3$