In the formula Z = l + left( frac{f_{1} - f_{ | vedclass

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(A) In the formula for the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} ight) 	imes c$,the term $c$ represents

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class length

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(A) In the formula for the mode of grouped data,$Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} ight) 	imes c$,the term $c$ represents the class size or class length of the modal class. The other terms are defined as follows: $l$ is the lower limit of the modal class,$f_{1}$ is the frequency of the modal class,$f_{0}$ is the frequency of the class preceding the modal class,and $f_{2}$ is the frequency of the class succeeding the modal class.

Class	$0-5$	$5-10$	$10-15$	$15-20$	$20-25$
Frequency	$10$	$15$	$12$	$20$	$9$

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Frequency	$12$	$16$	$8$	$6$	$8$

Income (in $Rs.$)	$1-200$	$201-400$	$401-600$	$601-800$
Number of employees	$14$	$15$	$14$	$7$

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

In the formula $Z = l + \left( \frac{f_{1} - f_{0}}{2f_{1} - f_{0} - f_{2}} \right) \times c$ for the mode,$c = \dots$

Similar Questions

For the following distribution:

Class $0-5$ $5-10$ $10-15$ $15-20$ $20-25$

Frequency $10$ $15$ $12$ $20$ $9$

the sum of the lower limits of the median class and the modal class is:

Calculate the mean of the following frequency distribution:

Class $10-20$ $20-30$ $30-40$ $40-50$ $50-60$

Frequency $12$ $16$ $8$ $6$ $8$

The daily income of a sample of $50$ employees is tabulated as follows:

Income (in $Rs.$) $1-200$ $201-400$ $401-600$ $601-800$

Number of employees $14$ $15$ $14$ $7$

Find the mean daily income of the employees. (in $.5$)

Find the mode of the following frequency distribution:

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

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

If $Z - M = 4$,then $M - \bar{x} = \dots$

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