Find the mode of the following frequency dist | vedclass

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(A) To find the mode of a grouped frequency distribution,we use the formula:$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$$1$

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(A) To find the mode of a grouped frequency distribution,we use the formula:$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$$1$. Identify the modal class: The class with the highest frequency is $40-60$ (frequency $26$).$2$. Here,the lower limit of the modal class $l = 40$,the size of the class interval $h = 20$.$3$. The frequency of the modal class $f_1 = 26$.$4$. The frequency of the class preceding the modal class $f_0 = 9$.$5$. The frequency of the class succeeding the modal class $f_2 = 23$.$6$. Substituting these values into the formula:$Mode = 40 + \left( \frac{26 - 9}{2(26) - 9 - 23} ight) 	imes 20$$Mode = 40 + \left( \frac{17}{52 - 32} ight) 	imes 20$$Mode = 40 + \left( \frac{17}{20} ight) 	imes 20$$Mode = 40 + 17 = 57$Thus,the mode is $57$.

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$	$100-120$
Frequency	$8$	$9$	$26$	$23$	$20$	$14$

Class	$0-20, 20-40, 40-60, 60-80, 80-100, 100-120$
Frequency	$10, f, 7, 6, 5, 4$

Class	$4-7$	$8-11$	$12-15$	$16-19$
Frequency	$5$	$4$	$9$	$10$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$2$	$4$	$10$	$20$	$18$	$20$	$16$	$10$

Find the mode of the following frequency distribution:

Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$

Frequency $8$ $9$ $26$ $23$ $20$ $14$

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Find the mode of the following frequency distribution: Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$ $100-120$ Frequency $8$ $9$ $26$ $23$ $20$ $14$