Find the mode of the following frequency dist | vedclass

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(D) 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$

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$64$

Vedclass · Answer

(D) 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$$1$. Identify the modal class (the class with the highest frequency). Here,the highest frequency is $55$,which corresponds to the class interval $60-70$.$2$. From the modal class:Lower limit $(l)$ = $60$Frequency of the modal class $(f_1)$ = $55$Frequency of the class preceding the modal class $(f_0)$ = $43$Frequency of the class succeeding the modal class $(f_2)$ = $37$Class size $(h)$ = $70 - 60 = 10$$3$. Substitute the values into the formula:$Mode = 60 + \left( \frac{55 - 43}{2(55) - 43 - 37} ight) 	imes 10$$Mode = 60 + \left( \frac{12}{110 - 80} ight) 	imes 10$$Mode = 60 + \left( \frac{12}{30} ight) 	imes 10$$Mode = 60 + \left( \frac{12}{3} ight)$$Mode = 60 + 4 = 64$Therefore,the mode of the given frequency distribution is $64$.

Class	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$
Frequency	$8$	$12$	$27$	$43$	$55$	$37$	$18$

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$
Frequency	$12$	$30$	$x$	$65$	$y$	$25$	$18$

Marks obtained	Number of students
More than or equal to $0$	$63$
More than or equal to $10$	$58$
More than or equal to $20$	$55$
More than or equal to $30$	$51$
More than or equal to $40$	$48$
More than or equal to $50$	$42$

Find the mode of the following frequency distribution:

Class $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$

Frequency $8$ $12$ $27$ $43$ $55$ $37$ $18$

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Find the mode of the following frequency distribution: Class $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ Frequency $8$ $12$ $27$ $43$ $55$ $37$ $18$