Find the mean of the following frequency distribution using step-deviation method Class 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 Frequency 7 10 15 8 10

Let us choose a = 25 , h = 10 , then d i = x i − 25 and u i = x i − 25 10 Using step-deviation method, the given data is shown as follows: Class Frequency ' ( f i ) ' Class mark ' ( x i ) ' ' d i = x i ' – 25 ' u i = ( x i − 25 ) 10 ' ' ( f i u i ) ' 0 – 10 7 5 - 20 - 2 - 14 10 – 20 10 15 - 10 - 1 - 10 20 – 30 15 25 0 0 0 30 – 40 8 35 10 1 8 40 – 50 10 45 20 2 20 Total ' Ʃ f i ' = 50 ' Ʃ f i u i = 4 ' The mean of the data is given by, x = a + ∑ f i u i ∑ f i × h = 25 + 4 50 × 10 = 25 + 4 5 = 125 + 4 5 = 129 5 = 25 . 8 Thus, the mean is 25 . 8 .

Find the mean of the following data, using step-deviation method: Class 5 – 15 15 - 20 20 - 35 35 - 45 45 - 55 55 - 65 65 - 75 Frequency 6 10 16 15 24 8 7

Let us choose a = 40 , h = 10 , then d i = x i − 40 and u i = x i − 40 10 Using step-deviation method, the given data is shown as follows: Class Frequency ' ( f i ) ' Class mark ' ( x i ) ' ' d i = x i ' – 40 ' u i = ( x i − 40 ) 10 ' ' ( f i u i ) ' 5 – 15 6 10 - 30 - 3 - 18 15 – 25 10 20 - 20 - 2 - 20 25 – 35 16 30 - 10 - 1 - 16 35 – 45 15 40 0 0 0 45 – 55 24 50 10 1 24 55 – 65 8 60 20 2 21 65 – 75 7 70 30 3 21 Total ∑ f i = 86 ∑ f i u i = 7 The mean of the data is given by, x = a + ∑ f i u i ∑ f i = 40 + 7 86 × 10 = 40 + 70 86 = 40 + 0 . 81 = 40 . 81 Thus, mean is 40 . 81

Find the mean of the following frequency distribution table using a suitable method: Class 20 – 30 30 – 40 40 – 50 50 – 60 60 - 70 Frequency 25 40 42 33 10

Let us choose a = 45 , h = 10 , then d i = x i − 45 and u i = x i − 45 10 Using step-deviation method, the given data is shown as follows: Weight Number of packets ' ( f i ) ' Class mark ' ( x i ) ' ' d i = x i ' – 45 ' u i = ( x i − 45 ) 10 ' ' ( f i u i ) ' 20 – 30 25 35 - 20 - 2 - 50 30 – 40 40 35 - 10 - 1 - 40 40 – 50 42 45 0 0 0 50 – 60 33 55 10 1 33 60 – 70 10 65 20 2 20 Total ' Ʃ f i ' = 150 ' Ʃ f i u i = ' - 37 The mean of the given data is given by, x = a + ∑ f i u i ∑ f i ' = 45 - 37 150 × 10 = 45 - 37 15 = 45 - 2 . 466 = 42 . 534 Hence, the mean is 42 . 534 .

In an annual examination, marks (out of 90 ) obtained by students of Class X in mathematics are given below: Marks Obtained 0 – 15 15 – 30 30 - 45 45 – 60 60 – 75 75 – 90 Number of students 2 4 5 20 9 10 Find the mean marks.

Let us choose a = 52 . 5 , h = 15 , then ' d i = x i – 52 . 5 and u i = ( x i − 52 . 5 ) 15 ' Using step-deviation method, the given data is shown as follows: Weight Number of students ' ( f i ) ' Class mark ' ( x i ) ' ' d i = x i – ' 37 . 5 ' u i = ( x i − 52 . 5 ) 15 ' ' ( f i u i ) ' 0 – 15 2 7 . 5 - 45 - 3 - 6 15 – 30 4 22 . 5 - 30 - 2 - 8 30 – 45 5 37 . 5 - 15 - 1 - 5 45 – 60 20 52 . 5 0 0 0 60 – 75 9 67 . 5 15 1 9 75 – 90 10 82 . 5 30 2 20 Total ∑ f i = 50 ∑ f i u i = 10 The mean of the given data is given by, x = a + ∑ i f i u i ∑ f i × h = 52 . 5 + 10 50 × 15 = 52 . 5 + 3 = 55 . 5 Thus, the mean is 55 . 5 .

Find the arithmetic mean of the following frequency distribution using step-deviation method: Age (in years) 18 - 24 24 - 30 30 - 36 36 - 42 42 - 48 48 - 54 Number of workers 6 8 12 8 4 2

Class Frequency ' ( f i ) Mid values ' ( x i ) ' ' u i = ( ( x i − A ) ) / ℎ = ( ( x i − 33 ) ) / 6 ' ' ( f i × u i ) ' 18 – 24 6 21 - 2 - 12 24 – 30 8 27 - 1 - 8 30 – 36 12 33 = A 0 0 36 – 42 8 39 1 8 42 – 48 4 45 2 8 48 – 54 2 51 3 6 Total ∑ f i = 40 ∑ f i × u i = 2 Now, A = 33 , h = 6 , ∑ f i = 40 and ∑ ( f i × u i ) = 2 ∴ Mean, x = A + h × ∑ f i u i ∑ f i = 33 + 6 × 2 40 = 33 + 0 . 3 = 33 . 3 ∴ x = 33 . 3 years

Find the mean of the following data using step-deviation method: Class 500 – 520 520 – 540 540 – 560 560 – 580 580 – 600 600 – 620 Frequency 14 9 5 4 3 5

Class Frequency ' ( f i ) ' Mid values ' ( x i ) ' ' u i = ( ( x i − A ) ) / ℎ = ( ( x i − 550 ) ) / 20 ' ' ( f i × u i ) ' 500 – 520 14 510 - 2 - 28 520 – 540 9 530 - 1 - 9 540 – 560 5 550 = A 0 0 560 – 580 4 570 1 4 580 – 600 3 590 2 6 600 – 620 5 610 3 15 ∑ f i = 40 ∑ f i × u i = - 12 Now, A = 550 , h = 20 , ∑ f i = 40 and ∑ ( f i × u i ) = - 12 ∴ Mean, x = A + h × ∑ f i × u i ∑ f i = 550 + 20 × - 12 40 = 550 - 6 = 544 ∴ x = 544

Find the mean age from the following frequency distribution: Age (in years) 25 – 29 30 – 34 35 – 39 40 – 44 45 – 49 50 – 54 55 – 59 Number of persons 4 14 22 16 6 5 3

Class Frequency ' ( f i ) ' Mid values ' ( x i ) ' ' u i = ( ( x i − A ) ) / ℎ = ( ( x i − 42 ) ) / 5 ' ' ( f i × u i ) ' 24 . 5 - 29 . 5 4 27 - 3 - 12 29 . 5 - 34 . 5 14 32 - 2 - 28 34 . 5 – 39 . 5 22 37 - 1 - 22 39 . 5 – 44 . 5 16 42 = A 0 0 44 . 5 – 49 . 5 6 47 1 6 49 . 5 – 54 . 5 5 52 2 10 54 . 5 – 59 . 5 3 57 3 9 ' Ʃ f i = 70 ' ' Ʃ ( f i × u i ) = - 37 ' Now, A = 42 , h = 5 , ∑ f i = 70 a n d ∑ ( f i × u i ) = - 37 ∴ Mean, X = A + h × ∑ ( f i × u i ) ∑ f i = 42 + 5 × - 37 70 = 42 - 2 . 64 = 39 . 36 ∴ X = 39 . 36 ∴ Mean age = 39 . 36 years .

Weight of 60 eggs were recorded as given below: Weight (in grams) 75 – 79 80 – 84 85 – 89 90 – 94 95 – 99 100 - 104 105 - 109 No. of eggs 4 9 13 17 12 3 2

Let us choose a = 92 , h = 5 , then ` d i = x i – 92 and u i = ( x i − 92 ) / 5 ' Using step-deviation method, the given data is shown as follows: Weight (in grams) Number of eggs ' ( f i ) ' Class mark ' ( x i ) ' ' d i = x i ' – 92 ' u i = ( x i − 92 ) / 5 ' ' ( f i u i ) ' 74 . 5 – 79 . 5 4 77 - 15 - 3 - 12 79 . 5 – 84 . 5 9 82 - 10 - 2 - 18 84 . 5 – 89 . 5 13 87 - 5 - 1 - 13 89 . 5 – 94 . 5 17 92 0 0 0 94 . 5 – 99 . 5 12 97 5 1 12 99 . 5 – 104 . 5 3 102 10 2 6 104 . 5 – 109 . 5 2 107 15 3 6 Total ∑ f i = 60 ∑ f i u i = - 19 The mean of the given data is given by, x = a + ∑ f i u i ∑ f i × h = 92 + - 19 60 × 5 = 92 – 1 . 58 = 90 . 42 ≈ 90 Thus, the mean weight to the nearest gram is 90 g .

R S Aggarwal and Veena Aggarwal Solutions for Exercise 1: Exercise 18A

Author:R S Aggarwal & Veena Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 1: Exercise 18A

Attempt the practice questions from Exercise 1: Exercise 18A 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 1: Exercise 18A 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 18A>Q 19

Find the mean of the following frequency distribution using step-deviation method

Class	$0 - 10$	$10 - 20$	$20 - 30$	$20 - 30$	$15$
Frequency	$0$	$0$	$15$	$8$	$10$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18A>Q 20

Find the mean of the following data, using step-deviation method:

Class	$2$	$15 - 20$	$20 - 35$	$35 - 45$	$45 - 55$	$55 - 65$	$65 - 75$
Frequency	$6$	$10$	$16$	$15$	$24$	$8$	$7$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18A>Q 22

Find the mean of the following frequency distribution table using a suitable method:

Class	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$
Frequency	$25$	$40$	$42$	$33$	$10$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18A>Q 23

In an annual examination, marks (out of $90$ ) obtained by students of Class X in mathematics are given below:

Marks Obtained	$0 - 15$	$15 - 30$	$30 - 45$	$45 - 60$	$60 - 75$	$75 - 90$
Number of students	$2$	$4$	$5$	$20$	$9$	$10$

Find the mean marks.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18A>Q 24

Find the arithmetic mean of the following frequency distribution using step-deviation method:

Age (in years)	$18 - 24$	$24 - 30$	$30 - 36$	$36 - 42$	$42 - 48$	$48 - 54$
Number of workers	$6$	$8$	$12$	$8$	$4$	$2$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18A>Q 25

Find the mean of the following data using step-deviation method:

Class	$500 - 520$	$520 - 540$	$540 - 560$	$560 - 580$	$580 - 600$	$600 - 620$
Frequency	$14$	$9$	$5$	$4$	$3$	$5$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18A>Q 26

Find the mean age from the following frequency distribution:

Age (in years)	$25 - 29$	$30 - 34$	$35 - 39$	$40 - 44$	$45 - 49$	$50 - 54$	$55 - 59$
Number of persons	$4$	$14$	$22$	$16$	$6$	$5$	$3$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18A>Q 28

Weight of $60$ eggs were recorded as given below:

Weight (in grams)	$75 - 79$	$80 - 84$	$85 - 89$	$90 - 94$	$95 - 99$	$100 - 104$	$105 - 109$
No. of eggs	$4$	$9$	$13$	$17$	$12$	$3$	$2$

R S Aggarwal and Veena Aggarwal Solutions for Exercise 1: Exercise 18A

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 1: Exercise 18A

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 1: Exercise 18A with Hints & Solutions

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