Compute variance and standard deviation for the following data: X 2 4 6 8 10 12 14 16 18 20 F 8 10 10 7 6 4 3 4 2 6

Given data set: X 2 4 6 8 10 12 14 16 18 20 F 8 10 10 7 6 4 3 4 2 6 Let the assumed mean be 10 . And let, u = X - 10 2 The other values for the given set of data as follows: X u F F u F u 2 2 - 4 8 - 32 128 4 - 3 10 - 30 90 6 - 2 10 - 20 40 8 - 1 7 - 7 7 10 0 6 0 0 12 1 4 4 4 14 2 3 6 12 16 3 4 12 36 18 4 2 8 32 20 5 6 30 150 Here, u ¯ = ∑ F u N = 29 60 Now the variance = σ u 2 = ∑ F u 2 N - u ¯ 2 = 499 60 - 29 60 2 ⇒ σ u 2 = 8 . 31 - 0 . 23 = 8 . 08 . Therefore, σ u = 8 . 08 = 2 . 84 . Now here, σ x = 2 × σ u = 2 × 2 . 84 = 5 . 68 Therefore, the value of the standard deviation is 5 . 68 .

Compute the variance and S.D. x 31 32 33 34 35 36 37 Frequency 15 12 10 8 9 10 6

Given Data: x 31 32 33 34 35 36 37 Frequency 15 12 10 8 9 10 6 The other values for the given set of data is as follows: X F F X F X 2 31 15 465 14415 32 12 384 12288 33 10 330 10890 34 8 272 9248 35 9 315 11025 36 10 360 12960 37 6 222 8214 Here, N = ∑ F = 70 , ∑ F i X i = 2348 a n d ∑ F i X i 2 = 79040 Therefore, the mean of the given data = x ¯ = ∑ f n N = 2348 70 = 33 . 54 Now the variance is given by, σ 2 = ∑ Fx 2 N - x ¯ 2 ⇒ 79040 70 - ( 33 . 5 ) 2 = 1129 . 14 - 1122 . 25 = 6 . 89 Therefore, the standard deviation is given by σ = 6 . 89 = 2 . 62

Following data gives age of 100 students in a college. Calculate variance and S.D. Age(In year) 16 17 18 19 20 21 No. of student 20 7 11 17 30 15

Given, the following data given an age of 100 students in a college: Age(In year) 16 17 18 19 20 21 No. of student 20 7 11 17 30 15 We need to find the variance and Standard Deviation for the following data x f xf x 2 f 16 20 320 5120 17 7 119 2023 18 11 198 3564 19 17 323 6137 20 30 600 12000 21 15 315 66 . 5 total 100 1875 35459 We get. x ¯ = Σ x f N = 1875 100 = 18 . 75 Now, σ = ∑ x 2 f N - x ¯ 2 = 35459 100 - ( 18 · 75 ) 2 = 354 . 59 - 351 . 56 = 3 . 03 Therefore , σ = 3 . 03 = 1 . 74

Find mean, variance and S.D. of the following data. Classes 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Freq. 7 14 6 13 9 15 11 10 15

let u = x - 15 . 5 10 class mid-value x i f i u i f i u i f i u i 2 10 - 20 15 . 5 7 0 0 0 20 - 30 20 . 5 14 0 . 5 7 3 . 5 30 - 40 25 . 5 6 1 6 6 40 - 50 30 . 5 13 1 . 5 19 . 5 29 . 25 50 - 60 35 . 5 9 2 18 36 60 - 70 40 . 5 15 2 . 5 37 . 5 93 . 75 70 - 80 45 . 5 11 3 33 99 80 - 90 50 . 5 10 3 . 5 35 122 . 5 90 - 100 55 . 5 15 4 60 240 100 216 630 now, ( u ¯ ) = ∑ f i u i N = 216 100 = 2 . 16 Variance is given by, V a r ( u ) = σ u 2 = ∑ f u 2 N - u ¯ 2 = 630 100 - 2 . 16 2 6 . 3 - 4 . 66 = 1 . 64 Thus Var ( μ ) = 1 . 64 Var ( x ) = h 2 · [ Var ( u ) ] = 10 × 1 . 64 = 16 · 4 So the standard deviation is, ∴ σ α = Var ( α ) = 16 · 4 = 4 . 04 Therefore, Variance and standard deviation of the given data is 16 . 4 and 4 . 04 .