Find the standard deviation for the following distribution: x 4 . 5 14 . 5 24 . 5 34 . 5 44 . 5 54 . 5 64 . 5 f 1 5 12 22 17 9 4

Var ( X ) = h 2 1 N ∑ i = 1 n f i u i 2 - 1 N ∑ i = 1 n f i u i 2 We have, N = 70 , ∑ i = 1 n f i u i = 22 , ∑ i = 1 n f i u i 2 = 130 , h = 10 Plugging all the values in the formula of variance: Var ( X ) = 10 2 1 70 × 130 - 1 70 × 22 2 = 100 130 70 - 22 70 2 = 100 13 7 - 121 1225 = 100 [ 1 . 857 - 0 . 0987 ] = 100 [ 1 . 7583 ] = 175 . 83 Standard Deviation, SD = Var ( X ) SD = Var ( X ) = 175 . 83 = 13 . 26

Table below shows the frequency f with which ' x ' alpha particles were radiated from a diskette x; 0 1 2 3 4 5 6 7 8 9 10 11 12 f: 51 203 383 525 532 408 273 139 43 27 10 4 2 Calculate the mean and variance7

Mean, X ¯ = ∑ f i x i ∑ f i = 10078 2600 = 3 . 88 x i f i f i x i x i - X ¯ x i - X ¯ ) 2 f i x i - X ¯ ) 2 0 51 0 - 3 . 88 15 . 05 767 . 55 1 203 203 - 2 . 88 8 . 29 1682 . 87 2 383 766 - 1 . 88 3 . 53 1351 . 99 3 525 1575 - 0 . 88 0 . 77 404 . 25 4 532 2128 0 . 12 0 . 014 7 . 448 5 408 2040 1 . 12 1 . 25 510 6 273 1638 2 . 12 4 . 49 1225 . 77 7 139 973 3 . 12 9 . 73 1352 . 47 8 43 344 4 . 12 16 . 97 729 . 71 9 27 243 5 . 12 26 . 21 707 . 67 10 10 100 6 . 12 37 . 45 374 . 5 11 4 44 7 . 12 50 . 69 202 . 76 12 2 24 8 . 12 65 . 93 131 . 86 ∑ f i ∑ f i f i x i x i - X ¯ = N = 10078 = 2600 = 9448 . 848 Variance, σ 2 = ∑ f i x i - X ¯ 2 N = 9448 . 848 2600 = 3 . 63 .

Find the mean, and standard deviation for the following data: Year render : 10 20 30 40 50 60 No . of persons ( cumulative ) : 15 32 51 78 97 109

Year render No. of persons (cumulative) f i u i = x i - 35 10 f i u i u i 2 f i u i 2 10 15 15 - 2 . 5 - 37 . 5 6 . 25 93 . 75 20 32 17 - 1 . 5 - 25 . 5 2 . 25 38 . 25 30 51 19 - 0 . 5 - 9 . 5 0 . 25 4 . 75 40 78 27 0 . 5 13 . 5 0 . 25 6 . 75 50 97 19 1 . 5 28 . 5 2 . 25 42 . 75 60 109 12 2 . 5 30 6 . 25 75 ∑ f i = N = 109 ∑ f i u i = - 0 . 5 ∑ f i u i 2 = 261 . 25 X ¯ = A + h ∑ f i u i N = 35 + 10 - 0 . 5 109 = 34 . 96 years σ 2 = h 2 ∑ f i u i 2 N - ∑ f i u i N 2 = 100 261 . 25 109 - 0 . 25 11881 = 100 × 2 . 396 = 239 . 6 σ = 239 . 6 = 15 . 47 years

Find the mean, and standard deviation for the following data: Marks: 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Frequency: 1 6 6 8 8 2 2 3 0 2 1 0 0 0 1

x i f i f i x i d i = x i - x f i d i f i d i 2 2 1 2 - 4 - 4 16 3 6 18 - 3 - 18 54 4 6 24 - 2 - 12 24 5 8 40 - 1 - 8 8 6 8 48 0 0 0 7 2 14 1 2 2 8 2 16 2 4 8 9 3 27 3 9 27 10 0 0 4 0 0 11 2 22 5 10 50 12 1 12 6 6 36 13 0 0 7 0 0 14 0 0 8 0 0 15 0 0 9 0 0 16 1 16 10 10 100 N = 40 ∑ f i x i = 239 ∑ f i d i = - 1 ∑ f i d i 2 = 325 Mean X = ∑ i = 1 n f i x i N = 239 40 = 5 . 9 = 6 Standard deviation ( σ ) = ∑ f i d i 2 N - ∑ f i d i N 2 = 325 40 - - 1 40 2 = 8 . 125 - 0 . 000625 = 8 . 124375 = 2 . 85 Here the required mean = 6 and standard deviation = 2 . 85

Find the standard deviation for the following data: x : 3 8 13 18 23 f : 7 10 15 10 6

x i f i f i x i f i x i 2 3 7 21 63 8 10 80 640 13 15 195 2535 18 10 180 3240 23 6 138 3174 N = 48 ∑ f i x i = 614 ∑ f i x i 2 = 9652 Standard deviation ( σ ) = ∑ f i x i 2 N - ∑ f i x i N 2 SD, σ = 201 . 08 - ( 12 . 79 ) 2 = 201 . 08 - 163 . 58 = 37 . 49 = 6 . 12

Find the standard deviation for the following data: x : 2 3 4 5 6 7 f : 4 9 16 14 11 6

x i f i f i x i f i x i 2 2 4 8 16 3 9 27 81 4 16 64 256 5 14 70 350 6 11 66 396 7 6 42 294 N = 60 ∑ f i x i = 277 ∑ f i x i 2 = 1393 S tan d a r d d e v i a t i o n ( σ ) = ∑ f i x i 2 N - ∑ f i x i N 2 = 1393 60 - 277 60 2 = 23 . 21 - ( 4 . 62 ) 2 = 23 . 21 - 21 . 34 = 1 . 87 = 1 . 37

E M B I B E

MATHEMATICS for CLASS XI>Chapter 32 - Statistics>EXERCISE 32.5

R D Sharma Solutions for Chapter: Statistics, Exercise 5: EXERCISE 32.5

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Statistics, Exercise 5: EXERCISE 32.5

Attempt the free practice questions on Chapter 32: Statistics, Exercise 5: EXERCISE 32.5 with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.