The mean of the following frequency distribut | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the missing frequency $f$,we use the step-deviation method for mean: $\bar{x} = A + \left( \frac{\sum f_i u_i}{\sum f_i} ight) 	imes c$

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(B) To find the missing frequency $f$,we use the step-deviation method for mean: $\bar{x} = A + \left( \frac{\sum f_i u_i}{\sum f_i} ight) 	imes c$.ClassFrequency $(f_i)$Midpoint $(x_i)$$u_i = \frac{x_i - A}{c}$$f_i u_i$$0-10$$8$$5$$-4$$-32$$10-20$$4$$15$$-3$$-12$$20-30$$20$$25$$-2$$-40$$30-40$$45$$35$$-1$$-45$$40-50$$64$$45=A$$0$$0$$50-60$$32$$55$$1$$32$$60-70$$f$$65$$2$$2f$$70-80$$8$$75$$3$$24$$80-90$$2$$85$$4$$8$$90-100$$2$$95$$5$$10$Total$\sum f_i = 185 + f$--$\sum f_i u_i = 2f - 55$Here,$A = 45$ and $c = 10$.Given $\bar{x} = 43.75$.$43.75 = 45 + \frac{2f - 55}{185 + f} 	imes 10$$-1.25 = \frac{10(2f - 55)}{185 + f}$$-1.25(185 + f) = 20f - 550$$-231.25 - 1.25f = 20f - 550$$550 - 231.25 = 20f + 1.25f$$318.75 = 21.25f$$f = \frac{318.75}{21.25} = 15$.Thus,the missing frequency $f = 15$.

Marks (out of $90$)	Number of candidates
More than or equal to $80$	$4$
More than or equal to $70$	$6$
More than or equal to $60$	$11$
More than or equal to $50$	$17$
More than or equal to $40$	$23$
More than or equal to $30$	$27$
More than or equal to $20$	$30$
More than or equal to $10$	$32$
More than or equal to $0$	$34$

Marks	$30-35$	$35-40$	$40-45$	$45-50$	$50-55$	$55-60$	$60-65$
Frequency	$14$	$16$	$18$	$23$	$18$	$8$	$3$

The mean of the following frequency distribution is $43.75$. Find the missing frequency $f$.
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$
Frequency $8$ $4$ $20$ $45$ $64$ $32$ $f$ $8$ $2$ $2$

Similar Questions

In the formula $M = l + \frac{(\frac{n}{2} - cf)}{f} \times h$ for the median,$f = \ldots \ldots \ldots$

The median of the following frequency distribution is $28.5$ and the total frequency is $60$. Find the missing frequencies $x$ and $y$.
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Frequency $5$ $x$ $20$ $15$ $y$ $5$

Find the median of the following frequency distribution:
Class $0-20$ $20-40$ $40-60$ $60-80$ $80-100$
Frequency $5$ $15$ $20$ $18$ $2$

The percentage of marks obtained by $100$ students in an examination are given below:

Marks $30-35$ $35-40$ $40-45$ $45-50$ $50-55$ $55-60$ $60-65$

Frequency $14$ $16$ $18$ $23$ $18$ $8$ $3$

Determine the median percentage of marks.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
Frequency	$8$	$4$	$20$	$45$	$64$	$32$	$f$	$8$	$2$	$2$

Class	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Frequency	$5$	$15$	$20$	$18$	$2$

The mean of the following frequency distribution is $43.75$. Find the missing frequency $f$.Class$0-10$$10-20$$20-30$$30-40$$40-50$$50-60$$60-70$$70-80$$80-90$$90-100$Frequency$8$$4$$20$$45$$64$$32$$f$$8$$2$$2$