Vedanta Optional Mathematics Teacher's Guide ~ 10 401 2. Compute standard deviation and its coefficients from the following method. i) direct method ii) short – cut method iii) step – deviation class interval 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Frequency 10 8 32 40 22 18 Solution i) To compute S.D. by direct method Class– interval Mid – value(m) Frequency(f) fm fm2 10 – 20 15 10 150 2250 20 – 30 25 8 200 5000 30 – 40 35 32 1120 39200 40 – 50 45 40 1800 8100 50 – 60 55 22 1210 66550 60 – 70 65 18 1170 76050 N = 130 ∑fm = 5650 270050 From table, N = 130, ∑fm = 5650, ∑fm2 = 270050, Solution Standard deviation ( ) = S.D.( )= ∑fm2 N – ∑fm N 2 = 270050 130 – 5650 130 2 = 2077.3076 – 1888.9053 = 188.4023 = 13.72 Mean(x) = ∑ fm N = 5650 130 = 43.46 Coefficient of S.D. = 6 x = 13.72 43.46 = 0.3157 ii) To compute standard deviation by short – cut method or assumed mean method. Let a = 45 Class– interval mid – value(m) frequency(f) d= m – a fm fd2 10 – 20 15 10 – 30 – 300 900 20 – 30 25 8 – 20 – 160 3200 30 – 40 35 32 – 10 – 320 3200 40 – 50 45 40 0 0 0 50 – 60 55 22 10 220 2200 60 – 70 65 18 20 360 7200 N = 130 – 200 24800
402 Vedanta Optional Mathematics Teacher's Guide ~ 10 Now, standard deviation is given by, = ∑fd2 N – ∑fd N 2 = 24800 130 – – 200 130 2 = 190.7652 – 2.3667 = 188.4025 = 13.72 Mean(x) = a + ∑ fm N = 40 + –200 130 = 45.46 coefficient of S.D. = 6 x = 13.72 43.46 = 0.3157 iii) To calculate standard deviation by step deviation method method Class– interval mid – value(m) frequency(f) d1 = m –a i fd1 fd1 2 10 – 20 15 10 – 3 – 30 90 20 – 30 25 8 – 20 – 16 32 30 – 40 35 32 – 1 – 32 32 40 – 50 45 40 0 0 0 50 – 60 55 22 1 22 22 60 – 70 65 18 2 36 72 N = 130 – 20 248 Standard deviation (S.D.) = = ∑fd1 N 2 – ∑fd1 N 2 × i = 248 130 – (–20) 130 2 × 10 = 1.9077 – 0.0237 × 10 = 1.88 × 10 = 13.72 Mean(x) = a + ∑ fd1 N × 10 = 40 + –20 130 × 10= 43.46 Coefficient of S.D. = 6 x = 13.72 43.46 = 0.3157
Vedanta Optional Mathematics Teacher's Guide ~ 10 403 3. Calculate the standard deviation and its coefficient of variation from the following. x 0 ≤ x ≤ 10 10 ≤ x ≤ 20 20 ≤ x ≤ 30 30 ≤ x ≤ 40 40 ≤ x ≤50 f 7 10 14 12 6 Solution Here, 0 ≤ x ≤ 10 means value of x is from 0 to 10 exclusively, we can write x belongs to class 0 – 10 etc. To calculate standard deviation of given data by using by step deviation method method Let a = 25, then d1 = m – a i = m – 25 i Class frequency(f) Mid – value (m) d1 = m –a i fd1 fd1 2 0 – 10 7 5 – 2 – 14 28 10 – 20 10 15 – 1 – 10 10 20 – 30 14 25 0 0 0 30 – 40 12 35 1 12 12 40 – 50 45 45 2 12 24 N = 49 0 70 Standard deviation = = ∑fd1 N 2 – ∑fd1 N 2 × i = 74 49 – 0 49 2 × 10 = 1.5210 × 10 = 12.33 Mean(x) = a + ∑ fd1 N × 10 = 25 + 0 49 × 10= 25 Coefficient of S.D. = 6 x = 12.33 25 = 0.4932 4. Calculate the standard deviation and its coefficient of variation from the following. (a) x less than 10 less than 20 thess than 30 less than 40 less than 50 f 12 19 24 33 40 Solution Given less than frequency table can be written in the following continuous class. Let i = 25, then d1 = m –a i
404 Vedanta Optional Mathematics Teacher's Guide ~ 10 Class Mid – value (m) frequency(f) d1 = m –a i fd1 fd1 2 0 – 10 5 12 – 2 – 24 48 10 – 20 15 19 – 12 = 7 – 1 – 7 7 20 – 30 25 24 – 19 = 5 0 0 0 30 – 40 35 33 – 24 = 9 1 9 9 40 – 50 45 40 – 33 = 7 2 14 28 N = 40 – 8 92 Standard deviation ( ) = ∑fd1 N 2 – ∑fd1 N 2 × i = 92 40 – –8 40 2 × 10 = 2.3 – 0.04 × 10 = 2.26 × 10 = 15.03 Mean(x) = a + ∑ fd1 N × i = 25 + – 8 40 × 10 = 23 Coefficient of variatin = 6 x × 100% = 15.03 23 = 65.35% b) x above 20 above 40 above 60 above 80 above 100 and less than 120 f 50 42 30 18 7 Solution Given more than cumulative frequency table can be written in the following continuous. Let a = 70, then d1 = m – a i =m – 70 10 Class Mid – value (m) frequency(f) d1 = m –a i fd1 fd1 2 20 – 40 30 50 – 48 = 8 – 2 – 16 32 40 – 60 50 12 – 1 – 12 12 60 – 80 70 18 0 0 0 80 – 100 90 11 1 11 11 100 – 120 110 7 2 14 28 N = 50 – 3 83 Standard deviation ( ) = ∑fd1 N 2 – ∑fd1 N 2 × i
Vedanta Optional Mathematics Teacher's Guide ~ 10 405 = 83 50 – 83 50 2 × 20 = 1.66 – 0.0036 × 20 = 1.6564 × 20 = 25.74 Mean(x) = a + ∑ fd1 N × i = 70 + – 3 50 × 20 = 68.8 Coefficient of variation (c.v) = 6 x × 100% = 25.74 68.8 × 100% = 37.41% c) Marks 0 – 10 0 – 20 0 – 30 0 – 40 0 –50 No. of students 7 18 30 42 50 Solution This is less than cumulative frequency table. We can change it as continuous frequency table as follows. Let a = 25, then d1 = m –a i = m – 25 10 Class frequency(f) Mid – value (m) d1 = m –a i fd1 fd1 2 0 – 10 7 5 – 2 – 14 28 10 – 20 18 – 7 =11 15 – 1 – 11 11 20 – 30 30 – 18 = 12 25 0 0 0 30 – 40 42 – 30 = 12 35 1 12 12 40 – 50 50 – 42 = 8 45 2 16 32 N = 50 0 3 83 Standard deviation = = ∑fd1 N 2 – ∑fd1 N 2 × i = 83 50 – 3 50 2 × 10 = 1.66 – 0.0036 × 10 = 12.87 Mean(x) = a + ∑ fd1 N × i = 25 + 3 50 × 10 = 25 + 0.6 = 25.6
406 Vedanta Optional Mathematics Teacher's Guide ~ 10 c.v = 6 x × 100% = 12.87 25.6 × 100 % = 50.27 % 5. From the given data which series is more variable (inconsistent). Variable 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Frequency section A 10 18 32 40 22 18 section B 18 22 40 32 20 10 Solution We calculate coefficient of variation two compare variability of two givens data. More c.v. more will be variability. class Series A B m fA dA 1 = m – 45 10 f AdB 1 f AdA 1 2 f B dB 1 = m – 45 10 f BdB 1 f BdB 1 10 – 20 15 10 – 3 – 30 90 18 – 3 – 54 162 20 – 30 25 18 – 2 – 36 72 22 – 2 – 44 88 30 – 40 35 32 – 1 – 32 32 40 – 1 – 40 40 40 – 50 45 40 0 0 0 32 0 0 0 50 – 60 55 22 1 22 22 20 1 20 21 60 – 70 65 18 2 36 72 10 2 20 40 NA = 140 – 40 288 142 = NB – 98 351 For series A, Standard deviation ( A )= ∑fAdA 1 N 2 – ∑fAdA 1 N 2 × i = 288 140 – –40 140 2 × 10 = 2.0571 – 0.0816 × 10 = 1.9755 × 10 = 14.05 Mean(x A)= a + ∑fAdA 1 NA × i = 45 + –40 140 2 × 10 = 42.14 c.v. (A) = 6A x A × 100%
Vedanta Optional Mathematics Teacher's Guide ~ 10 407 = 14.05 42.14 × 100 % = 33.34 % Again for series B, Standard deviation ( B )= ∑fBdB 1 NB 2 – ∑fBdB 1 NB 2 × i = 351 142 – –98 142 2 × 10 = 2.4718 – 0.4763 × 10 = 1.9955 × 10 = 1.413 Mean(x B)= a + ∑fBdB 1 NB × i = 45 + –98 142 2 × 10 = 38.09 c.v. (B) = 6B x B × 100% = 14.13 38.09 × 100 % = 37.01 % Since c.v. (B) > c.v. (A), the series B is more variable or more inconsistent. Questions for practice 1. If Q1 = 45, Q.D = 20, find the third quartile and coefficient of Quartile deviation. 2. If coefficient of Q.D is 0.5, third quartile is 25, find the first quartile. 3. Find the Quartile Deviation and its coefficient for the following given data. (a) x 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 f 2 4 6 12 15 12 6 4 3 (b) x less than 20 less than 30 less than 40 less than 50 f 5 10 20 25 (c) x more than 10 more than 20 more than 30 more than 40 f 40 30 20 10 5
408 Vedanta Optional Mathematics Teacher's Guide ~ 10 (d) Marks 0 –10 0 – 20 0 – 30 0 – 40 0 – 50 No. of students 5 15 25 40 50 (e) Marks 0 – 50 0 – 40 0 – 30 0 – 20 0 – 10 No. of students 40 30 20 10 5 4. Taking class interval 10 find the quartile deviation from the given data 40, 50, 60, 70, 50, 80, 70, 90, 13, 22 70, 80, 50, 60, 70, 85, 95, 65, 50, 45 22, 45, 60, 70, 85, 70, 90, 72, 55, 49 5. If Q1 = 32, find the value of x and then quartile deviation from given date. x 0 – 20 0 – 40 0 – 60 0 – 80 0 – 100 f 10 20 30 + x 40 + x 50 + x Mean Deviation 1. Find the coefficient of mean deviation of a continuous series having 20 samples whose mean is 40 and ∑f|d| = 240. 2. Find the coefficient of mean deviation from median whose median is 35 and number of item is 40 and ∑f|d| = 440. 3. Find the mean deviation from mean and its coefficients. (a) Marks 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 No. of students 4 10 7 6 3 (b) x 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50 f 5 13 25 40 50 4. Find the mean deviation from median and its coefficient. (a) x 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 y 5 10 15 10 5 (b) x 0 – 10 0 – 20 0 – 30 0 – 40 0 – 50 f 2 5 10 15 20
Vedanta Optional Mathematics Teacher's Guide ~ 10 409 (c) Marks 10 – 20 20 – 30 20 – 40 40 – 50 No. of students 5 7 15 3 Standard Deviation: 1. In a continuous series, N = 25, ∑fd = 480, ∑fd2 = 3240, d=(m – x ) find the coefficient of standard deviation. 2. In a continuous series N = 40, ∑fm = 100, ∑fd2 = 4060 find the coefficient of standard deviation. 2. In a continuous series N = 40, ∑fm = 100, ∑fd2 = 4060 find the coefficient of standard deviation. 3. In a continuous series N = 40, ∑fd1 = 7, ∑fd1 2 = 75, i = 10, assumed mean a = 20, find the standard deviation and its coefficient 4. Find the standard deviation and its coefficient from the following data : (a) x 10 – 20 20 – 30 20 – 40 40 – 50 50 – 60 y 10 15 20 10 5 (b) x 0 –6 0 – 12 0 – 18 0 – 24 0 – 30 y 5 10 15 20 25 (c) marks less than 10 less than 20 less than 30 less than 40 less than 50 No. of students 5 9 12 20 25 5. Prepare a frequency distribution table taking class interval 10. Calculate the standard deviation and its coefficient. 20, 22, 24, 25, 28, 10, 22, 70, 80, 45 33, 45, 37, 80, 75, 95, 80, 75, 78, 88 60, 66, 65, 68, 78, 90, 88, 78, 79, 90