Find the mean, by Direct Method: Class interval 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Frequency 11 15 20 30 14 10

If x 1 , x 2 , x 3 , . . . . . . , x n are the class marks with corresponding frequencies f 1 , f 2 , f 3 , . . . . . . . , f n respectively, then, Mean = x ¯ = f 1 x 1 + f 2 x 2 + f 3 x 3 + . . . . . . . + f n x n f 1 + f 2 + f 3 + . . . . . . . + f n = ∑ f i x i ∑ f i From the given data, we can plot the below table: Class interval Frequency f i Class Mark x i f i x i 10 - 20 11 10 + 20 2 = 30 2 = 15 165 20 - 30 15 25 375 30 - 40 20 35 700 40 - 50 30 45 1350 50 - 60 14 55 770 60 - 70 10 65 650 ∑ f i = 100 ∑ f i x i = 4010 Mean = ( x ¯ ) = Σ f i x i Σ f i = 4010 100 = 40 . 1 Hence, the mean is 40 . 1 .

Find the mean, by Direct Method: Class interval 25 - 35 35 - 55 45 - 55 55 - 65 65 - 75 Frequency 6 10 8 12 4

If x 1 , x 2 , x 3 , . . . . . . , x n are the class marks with corresponding frequencies f 1 , f 2 , f 3 , . . . . . . . , f n respectively, then, Mean = x ¯ = f 1 x 1 + f 2 x 2 + f 3 x 3 + . . . . . . . + f n x n f 1 + f 2 + f 3 + . . . . . . . + f n = ∑ f i x i ∑ f i From the given data, we can plot the below table: Class interval Frequency f i Class Mark x i f i x i 25 - 35 6 25 + 35 2 = 60 2 = 30 180 35 - 45 10 40 400 45 - 55 8 50 400 55 - 65 12 60 720 65 - 75 4 70 280 ∑ f i = 40 ∑ f i x i = 1980 Mean = ( x ¯ ) = Σ f i x i Σ f i = 1980 40 = 49 . 5 Hence, the mean is 49 . 5 . Note: Answer mentioned in the book is incorrect the correct answer is: 49 . 5 .

Find the mean, by Direct Method: Class interval 0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 Frequency 6 9 15 12 8

If x 1 , x 2 , x 3 , . . . . . . , x n are the class marks with corresponding frequencies f 1 , f 2 , f 3 , . . . . . . . , f n respectively, then, Mean = x ¯ = f 1 x 1 + f 2 x 2 + f 3 x 3 + . . . . . . . + f n x n f 1 + f 2 + f 3 + . . . . . . . + f n = ∑ f i x i ∑ f i From the given data, we can plot the below table: Class interval Frequency f i Class Mark x i f i x i 0 - 100 6 0 + 100 2 = 100 2 = 50 300 100 - 200 9 150 1350 200 - 300 15 250 3750 300 - 400 12 350 4200 400 - 500 8 450 3600 ∑ f i = 50 ∑ f i x i = 13200 Mean = ( x ¯ ) = Σ f i x i Σ f i = 13200 50 = 264 Hence, the mean is 264 .

Find the mean by Short-cut Method (Assumed Mean method). Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 12 16 6 7 9

From the given data, we can plot the below table: Class interval Frequency f i Class Mark x i Deviations d i = x i - 25 f i d i 0 - 10 12 0 + 10 2 = 10 2 = 5 5 - 25 = - 20 - 240 10 - 20 16 15 - 10 - 160 20 - 30 6 25 = a 0 0 30 - 40 7 35 10 70 40 - 50 9 45 20 180 ∑ f i = 50 ∑ f i d i = - 150 Let a = 25 be the assumed mean, then we have, Mean ( x ¯ ) = a + Σ f i d i Σ f i = 25 + ( - 150 ) 50 = 25 – 3 = 22 Hence, the mean is 22 . Note: Answer mentioned in the book is incorrect the correct answer is: 22 .

Find the mean by Short-cut Method (Assumed Mean method). Class interval 100 - 120 120 - 140 140 - 160 160 - 180 180 - 200 Frequency 10 20 30 15 5

From the given data, we can plot the below table: Class interval Frequency f i Class Mark x i Deviations d i = x i - a f i d i 100 - 120 10 100 + 120 2 = 220 2 = 110 110 - 150 = - 40 - 400 120 - 140 20 130 - 20 - 400 140 - 160 30 150 = a 0 0 160 - 180 15 170 20 300 180 - 200 5 190 40 200 ∑ f i = 80 ∑ f i d i = - 300 Let, a = 150 be the assumed mean, then we have, Mean ( x ¯ ) = a + Σ f i d i Σ f i = 150 - 300 80 = 150 – 3 . 75 = 146 . 25 Hence, the mean is 146 . 25 .

Find the mean, by Step Deviation Method. Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Number of students 12 18 27 20 17 6

From the given data we can plot the below table: Class interval Frequency f i Class Mark x i Deviations d i = x i - 25 u i = d i h = x i - a 10 f i u i 0 - 10 12 0 + 10 2 = 10 2 = 5 5 - 25 = - 20 - 20 10 = - 2 - 24 10 - 20 18 15 - 10 - 1 - 18 20 - 30 27 25 = a 0 0 0 30 - 40 20 35 10 1 20 40 - 50 17 45 20 2 34 50 - 60 6 55 30 3 18 ∑ f i = 100 ∑ f i u i = 30 Let, the assumed mean a = 25 and class size h = 10 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 25 + 30 100 × 10 = 25 + 3 = 28 Hence, the mean is 28 .

Find the mean, by Step Deviation Method. Class interval 0 - 20 0 - 40 40 - 60 60 - 80 80 - 100 Frequency 15 18 21 29 17

From the given data, we can plot the below table: Class interval Frequency f i Class Mark x i Deviations d i = x i - 50 u i = d i h = x i - 50 20 f i u i 0 - 20 15 0 + 20 2 = 20 2 = 10 10 - 50 = - 40 - 40 20 = - 2 - 30 20 - 40 18 30 - 20 - 1 - 18 40 - 60 21 50 = a 0 0 0 60 - 80 29 70 20 1 29 80 - 100 17 90 40 2 34 ∑ f i = 100 ∑ f i u i = 15 Let, the assumed mean a = 50 and class size h = 20 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 50 + 15 100 × 20 = 50 + 3 = 53 Hence, the mean is 53 .

Find the mean, by Step Deviation Method. Class interval 50 - 70 70 - 90 90 - 110 110 - 130 130 - 150 150 - 170 Frequency 18 12 13 27 8 22

From the given data, we can plot the below table: Class interval Frequency f i Class Mark x i Deviations d i = x i - 100 u i = d i h = x i - a 20 f i u i 50 - 70 18 50 + 70 2 = 120 2 = 60 60 - 100 = - 40 - 40 20 = - 2 - 36 70 - 90 12 80 - 20 - 1 - 12 90 - 110 13 100 = a 0 0 0 110 - 130 27 120 20 1 27 130 - 150 8 140 40 2 16 150 - 170 22 160 60 3 66 ∑ f i = 100 ∑ f i u i = 61 Let, the assumed mean a = 100 and class size h = 20 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 100 + 61 100 × 20 = 100 + 12 . 2 = 112 . 2 Hence, the mean is 112 . 2 .

E M B I B E

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE

J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 1: EXERCISE

Author:J P Mohindru & Bharat Mohindru

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 1: EXERCISE

Attempt the free practice questions on Chapter 14: Statistics, Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. Modern's abc+ of Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.