The following table shows the daily wages of workers in a factory: Daily wages ( in ₹ ) 0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 Number of workers 40 32 48 22 8 Find the median daily wage income of the workers.

Class Frequency f Cumulative Frequency c f 0 - 100 40 40 100 - 200 32 72 200 - 300 48 120 300 - 400 22 142 400 - 500 8 150 N = ∑ f = 150 Now, N = 150 ⇒ N 2 = 75 . The cumulative frequency just greater than 75 is 120 and the corresponding class is 200 - 300 . Thus, the median class is 200 - 300 . ∴ l = 200 , h = 100 , f = 48 , cumulative frequency of preceding class ( cf ) = 72 and N 2 = 75 . Now, median is given as, M = l + h × ( N 2 - cf ) f ⇒ 200 + 100 × ( 75 - 72 ) 48 ⇒ 200 + 6 . 25 Hence, the median daily wage income of the workers is ₹ 206 . 25 .

The median of the following data is 16 . Find the missing frequencies a and b if the total of frequencies is 70 . Class 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 35 - 40 Frequency 12 a 12 15 b 6 6 4

Class Frequency f Cumulative Frequency c f 0 - 5 12 12 5 - 10 a 12 + a 10 - 15 12 24 + a 15 - 20 15 39 + a 20 - 25 b 39 + a + b 25 - 30 6 45 + a + b 30 - 35 6 51 + a + b 35 - 40 4 55 + a + b N = ∑ f = 70 Let a and b be the missing frequencies of class intervals 5 - 10 and 20 - 25 respectively. Then, 55 + a + b = 70 ⇒ a + b = 15 . . . . . ( 1 ) Median is 16 , which lies in 15 - 20 . So, the median class is 15 - 20 . &there4; l = 15 , h = 5 , N = 70 , f = 15 and cumulative frequency ( cf ) = 24 + a . Now, median is given as, M = l + ( N 2 - cf ) f × h 16 = 15 + ( 35 - 24 - a ) 15 × 5 16 = 15 + ( 11 - a ) 3 16 - 15 = ( 11 - a ) 3 1 × 3 = 11 - a a = 11 - 3 a = 8 ∴ b = 15 - a [From ( 1 ) ] b = 15 - 8 b = 7 Hence, the missing frequency a = 8 & b = 7 .

If the median of the following frequency distribution is 32 . 5 , find the values of f 1 and f 2 . Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Total Frequency f 1 5 9 12 f 2 3 2 40

class Frequency f Cumulative Frequency c f 0 - 10 f 1 f 1 10 - 20 5 f 1 + 5 20 - 30 9 f 1 + 14 30 - 40 12 f 1 + 26 40 - 50 f 2 f 1 + f 2 + 26 50 - 60 3 f 1 + f 2 + 29 60 - 70 2 f 1 + f 2 + 31 N = ∑ f = 40 f 1 + f 2 + 31 = 40 ⇒ f 1 + f 2 = 9 . . . . . . . ( 1 ) Now, The median is 32 . 5 which lies in 30 - 40 . Hence, median class is 30 - 40 . ∴ l = 30 , N 2 = 20 , h = 10 , f = 12 , cumulative frequency of preceding class ( cf ) = 14 + f 1 . Now, median is given as, M = l + ( N 2 - cf ) f × h 32 . 5 = 30 + ( 20 - ( 14 + f 1 ) ) 12 × 10 2 . 5 = ( 60 - 10 f 1 ) 12 30 = 60 - 10 f 1 f 1 = 30 10 f 1 = 3 From equation ( 1 ) , we get f 2 = 9 - f 1 f 2 = 9 - 3 f 2 = 6 Hence, the value of f 1 = 3 & f 2 = 6 .

Calculate the median for the following data: Age (in years) 19 - 25 26 - 32 33 - 39 40 - 46 47 - 53 54 - 60 Frequency 35 96 68 102 35 4

First, we will convert the data into exclusive form. Class Frequency f Cumulative frequency c f 18 . 5 - 25 . 5 35 35 25 . 5 - 32 . 5 96 131 32 . 5 - 39 . 5 68 199 39 . 5 - 46 . 5 102 301 46 . 5 - 53 . 5 35 336 53 . 5 - 60 . 5 4 340 N = ∑ f = 340 ∴ l = 32 . 5 , h = 7 , f = 68 , cumulative frequency of preceding class ( cf ) = 131 and N 2 = 170 .Now, N = 340 ⇒ N 2 = 170 . The cumulative frequency just greater than 170 is 199 and the corresponding class is 32 . 5 - 39 . 5 . Thus, the median class is 32 . 5 - 39 . 5 . Now, median is given as, M = l + ( N 2 - cf ) f × h ⇒ 32 . 5 + ( 170 - 131 ) 68 × 7 ⇒ 32 . 5 + 4 . 01 ⇒ 36 . 51 Hence, the median is 36 . 51 . “Note: Answer mentioned in the book is incorrect the correct answer is: 36 . 51 ”

Find the median wages for the following frequency distribution: Wages per day (in Rs) 61 - 70 71 - 80 81 - 90 91 - 100 101 - 110 111 - 120 No. of women workers 5 15 20 30 20 8

We need to convert the class into exclusive form, so we get, Class Frequency f Cumulative Frequency c f 60 . 5 - 70 . 5 5 5 70 . 5 - 80 . 5 15 20 80 . 5 - 90 . 5 20 40 90 . 5 - 100 . 5 30 70 100 . 5 - 110 . 5 20 90 110 . 5 - 120 . 5 8 98 N = ∑ f = 98 Now, N = 98 ⇒ N 2 = 49 . The cumulative frequency just greater than 49 is 70 and the corresponding class is 90 . 5 - 100 . 5 . Thus, the median class is 90 . 5 - 100 . 5 . Now, l = 90 . 5 , h = 10 , f = 30 , cumulative frequency of preceding class ( cf ) = 40 and N 2 = 49 . Now, median is given as, M = l + h × ( N 2 - cf ) f ⇒ 90 . 5 + 10 × ( 49 - 40 ) 30 ⇒ 90 . 5 + 3 ⇒ 93 . 5 Hence, median wages is ₹ 93 . 5 .

Find the median from the following data: Class 1 - 5 6 - 10 11 - 15 16 - 20 21 - 25 26 - 30 31 - 35 36 - 40 41 - 45 Frequency 7 10 16 32 24 16 11 5 2

First we convert the given data into exclusive form, so we get: Class Frequency f Cummulative Frequency C f 0 . 5 - 5 . 5 7 7 5 . 5 - 10 . 5 10 17 10 . 5 - 15 . 5 16 33 15 . 5 - 20 . 5 32 65 20 . 5 - 25 . 5 24 89 25 . 5 - 30 . 5 16 105 30 . 5 - 35 . 5 11 116 35 . 5 - 40 . 5 5 121 40 . 5 - 45 . 5 2 123 N = ∑ f = 123 ∴ l = 15 . 5 , h = 5 , f = 32 cumulative frequency of the preceding class ( cf ) = 33 and N 2 = 61 . 5 .Now, N = 123 ⇒ N 2 = 61 . 5 . The cumulative frequency just greater than 61 . 5 is 65 and the corresponding class is 15 . 5 - 20 . 5 . Thus, the median class is 15 . 5 - 20 . 5 . Now, median is given as, M = l + ( N 2 - cf ) f × h ⇒ 15 . 5 + ( 61 . 5 - 33 ) 32 × 5 ⇒ 15 . 5 + 4 . 45 ⇒ 19 . 95 Hence, the median is 19 . 95 .

Find the median from the following data: Marks No. of students Below 10 12 Below 20 32 Below 30 57 Below 40 80 Below 50 92 Below 60 116 Below 70 164 Below 80 200

Class Cumulative frequency (cf) frequency (f) 0 - 10 12 12 10 - 20 32 20 20 - 30 57 25 30 - 40 80 23 40 - 50 92 12 50 - 60 116 24 60 - 70 164 48 70 - 80 200 36 N = ∑ f = 200 Now, N = 200 ⇒ N 2 = 100 . The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 - 60 . Thus, the median class is 50 - 60 . ∴ l = 50 , h = 10 , f = 24 , cumulative frequency of preceding class ( cf ) = 92 and N 2 = 100 . Now, median is given as, M = l + ( N 2 - cf ) f × h ⇒ 50 + ( 100 - 92 ) 24 × 10 ⇒ 50 + 3 . 33 ⇒ 53 . 33 Hence, the median is 53 . 33 .

R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: Exercise 18B

Author:R S Aggarwal & Veena Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: Exercise 18B

Attempt the free practice questions from Exercise 2: Exercise 18B with hints and solutions to strengthen your understanding. Secondary School Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: Exercise 18B with Hints & Solutions

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 3

The following table shows the daily wages of workers in a factory:

Daily wages $(in ₹)$	$0 - 100$	$100 - 200$	$200 - 300$	$300 - 400$	$400 - 500$
Number of workers	$40$	$32$	$48$	$22$	$8$

Find the median daily wage income of the workers.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 8

The median of the following data is $16$ . Find the missing frequencies $a$ and $b$ if the total of frequencies is $70$ .

Class	$0 - 5$	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$	$35 - 40$
Frequency	$12$	$a$	$12$	$15$	$b$	$6$	$6$	$4$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 9

In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies $x$ and $y$ .

Runs scored	$2500 - 3500$	$3500 - 4500$	$4500 - 5500$	$5500 - 6500$	$6500 - 7500$	$7500 - 8500$
Number of batsmen	$5$	$x$	$y$	$12$	$6$	$2$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 10

If the median of the following frequency distribution is $32.5$ , find the values of $f_{1}$ and $f_{2}$ .

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	Total
Frequency	$f_{1}$	$5$	$9$	$12$	$f_{2}$	$3$	$2$	$40$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 11

Calculate the median for the following data:

Age (in years)	$19 - 25$	$26 - 32$	$33 - 39$	$40 - 46$	$47 - 53$	$54 - 60$
Frequency	$35$	$96$	$68$	$102$	$35$	$4$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 12

Find the median wages for the following frequency distribution:

Wages per day (in Rs)	$61 - 70$	$71 - 80$	$81 - 90$	$91 - 100$	$101 - 110$	$111 - 120$
No. of women workers	$5$	$15$	$20$	$30$	$20$	$8$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 13

Find the median from the following data:

Class	$1 - 5$	$6 - 10$	$11 - 15$	$16 - 20$	$21 - 25$	$26 - 30$	$31 - 35$	$36 - 40$	$41 - 45$
Frequency	$7$	$10$	$16$	$32$	$24$	$16$	$11$	$5$	$2$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 14

Find the median from the following data:

Marks	No. of students
Below $10$	$12$
Below $20$	$32$
Below $30$	$57$
Below $40$	$80$
Below $50$	$92$
Below $60$	$116$
Below $70$	$164$
Below $80$	$200$

R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: Exercise 18B

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: Exercise 18B

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 2: Exercise 18B with Hints & Solutions

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