Find the mode of the following frequency dist | vedclass

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(B) The formula for the mode of a grouped frequency distribution is given by:$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$$1

Vedclass · Accepted Answer

$46$

Vedclass · Answer

(B) The formula for the mode of a grouped frequency distribution is given by:$Mode = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} ight) 	imes h$$1$. Identify the modal class: The class with the highest frequency is $40-50$,so the modal class is $40-50$.$2$. Identify the parameters:- Lower limit of the modal class $(l)$ = $40$- Frequency of the modal class $(f_1)$ = $42$- Frequency of the class preceding the modal class $(f_0)$ = $24$- Frequency of the class succeeding the modal class $(f_2)$ = $30$- Class size $(h)$ = $10$$3$. Substitute the values into the formula:$Mode = 40 + \left( \frac{42 - 24}{2(42) - 24 - 30} ight) 	imes 10$$Mode = 40 + \left( \frac{18}{84 - 54} ight) 	imes 10$$Mode = 40 + \left( \frac{18}{30} ight) 	imes 10$$Mode = 40 + \left( \frac{18}{3} ight) = 40 + 6 = 46$Thus,the mode is $46$.

Runs scored	Number of batsmen
$3000-4000$	$4$
$4000-5000$	$18$
$5000-6000$	$9$
$6000-7000$	$7$
$7000-8000$	$6$
$8000-9000$	$3$
$9000-10000$	$1$
$10000-11000$	$1$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
Frequency	$4$	$7$	$x$	$18$	$15$	$8$	$7$

Find the mode of the following frequency distribution:

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$

Frequency $6$ $8$ $17$ $24$ $42$ $30$ $15$ $8$

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The mode of the following frequency distribution is $37$. Find the missing frequency $x$.
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The mode of the observations $4, 5, 6, 3, 4, 3, 3, 2, 3, 5$ is $\ldots \ldots \ldots \ldots .$

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Find the mode of the following frequency distribution: Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ Frequency $6$ $8$ $17$ $24$ $42$ $30$ $15$ $8$