Find the mean by Short-cut Method (Assumed Mean method). Class interval 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 200 Frequency 20 35 52 44 38 31

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 0 - 20 20 0 + 20 2 = 20 2 = 10 10 - 50 = - 40 - 800 20 - 40 35 30 - 20 - 700 40 - 60 52 50 = a 0 0 60 - 80 44 70 20 880 80 - 100 38 90 40 1520 100 - 120 31 110 60 1860 ∑ f i = 220 ∑ f i d i = 2760 Let, a = 50 be the assumed mean, then we have, Mean ( x ¯ ) = a + Σ f i d i Σ f i = 50 + ( 2760 ) 220 = 50 + 12 . 55 = 62 . 55 approx . Hence, the mean is 62 . 55 .

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 .

Find the mean, by Step Deviation Method. Class interval 0 - 50 50 - 100 100 - 150 150 - 200 200 - 250 250 - 300 Frequency 17 35 43 40 21 24

From the given data, we can plot the below table: Class interval Frequency f i Class Mark x i Deviations d i = x i - 125 u i = d i h = x i - a 50 f i u i 0 - 50 17 0 + 50 2 = 50 2 = 25 25 - 125 = - 100 - 100 50 = - 2 - 34 50 - 100 35 75 - 50 - 1 - 35 100 - 150 43 125 = a 0 0 0 150 - 200 40 175 50 1 40 200 - 250 21 225 100 2 42 250 - 300 24 275 150 3 72 ∑ f i = 180 ∑ f i u i = 85 Let, the assumed mean a = 125 and class size h = 50 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 125 + 85 180 × 50 = 125 + 23 . 61 = 148 . 61 approx Hence, the mean is 148 . 61 .

Find the arithmetic mean of the following frequency distribution: Class interval 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59 Frequency 14 22 16 6 5 3 4

Here, the class interval is non-overlapping. ∴ Class size h = 29 - 25 + 30 - 29 = 4 + 1 = 5 Class interval Frequency f i Class Mark x i Deviations d i = x i - 42 u i = d i h = x i - 42 5 f i u i 25 - 29 14 25 + 29 2 = 54 2 = 27 27 - 42 = - 15 - 15 5 = - 3 - 42 30 - 34 22 32 - 10 - 2 - 44 35 - 39 16 37 - 5 - 1 - 16 40 - 44 6 42 = a 0 0 0 45 - 49 5 47 5 1 5 50 - 54 3 52 10 2 6 55 - 59 4 57 15 3 12 ∑ f i = 70 ∑ f i u i = - 79 Let, the assumed mean a = 42 and class size h = 5 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 42 + ( - 79 ) 70 × 5 = 42 - 79 14 = 42 - 5 . 64 = 36 . 36 approx Hence, the mean is 36 . 36 .

Find the mean age (in years) from the following frequency distribution: Age (in years) 15 - 19 20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 Frequency 3 13 21 15 5 4 2

Here, the class interval is non-overlapping. ∴ Class size h = 19 - 15 + 20 - 19 = 4 + 1 = 5 Age (in years) Frequency f i Class Mark x i Deviations d i = x i - 32 u i = d i h = x i - 32 5 f i u i 15 - 19 3 15 + 19 2 = 34 2 = 17 17 - 32 = - 15 - 15 5 = - 3 - 9 20 - 24 13 22 - 10 - 2 - 26 25 - 29 21 27 - 5 - 1 - 21 30 - 34 15 32 = a 0 0 0 35 - 39 5 37 5 1 5 40 - 44 4 42 10 2 8 45 - 49 2 47 15 3 6 ∑ f i = 63 ∑ f i u i = - 37 Let, the assumed mean a = 32 and class size h = 5 , then we have, Mean x ¯ = a + Σ f i u i Σ f i × h = 32 + ( - 37 ) 63 × 5 = 32 - 185 63 = 32 – 2 . 94 = 29 . 06 approx Hence, the mean age is 29 . 06 years.