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# Module 6

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### Module 6

1. 1. Unit – 6 SKEWNESS Introduction Measures of Central Tendency give us an estimate of the representative value of a series, the measures of dispersion on the other hand gives an indication of the extent to which the items scatter away from that representative value. But these two measures fail to give any idea as to the nature of distribution. It is quite incorrect to declare that the distributions are exactly alike in nature on the basis of the similarity of the mean and standard deviation. Thus they may be decidedly dissimilar in respect of their formation. To study the characteristics of a frequency distribution, it is necessary to know whether the distribution is normal or asymmetrical and if it is an asymmetrical distribution, to what extent it deviates from the normal curve, such measures of asymmetry are called measures of skewness. Meaning & Definition:- A frequency distribution is said to be symmetrical when the values of the variable equidistant from their mean have equal frequencies. If a frequency distribution is not symmetrical, it is said to be asymmetrical or skewed. Any deviation from symmetry is called skewness. Therefore, the term Skewness refers to lack of symmetry. That is when a distribution is not symmetrical, it is called a skewed distribution. A perfectly symmetrical distribution is one when its values are plotted on a graph, it represents a bell shaped curve of normal distribution This can be explained with the help of a graph Y - - Frequency - - - 50% 50% - 0=M=Z The above case is of symmetrical distribution. The hump is in the middle here and the right and left tails are of equal length. In the case of asymmetrical distribution one toil is longer than the other. Such distributions ae said to be skewed. The skewness again can be positive or negative. Positively skewed distribution can be shown as below diagrammatically. Y - - asymmetrical series - 0 >M>Z - - 0 MD M 0 variables X In the above diagram, the right tail is longer than the left tail. In such a case the skewness is said to be positive in the distribution, the mode come first and the median afterwards as we move along the x axis. 109
2. 2. Negatively Skewed Distribution Y -Frequency - - - - 0 0 M Z variable X In the above graph one can observe that the left tail is longer than the right tail. Such a distribution is negatively skewed. Here mean comes first as we go along the x axis and mode comes last. Definition:- Having understood the meaning of skewness, consider the following definitions. “ Skewness or asymmetry is the attribute of a frequency distribution that extends further on one side of the class with the highest frequency that on the other” Simpson & Kajka “ When a series is not symmetrical, it is said to be asymmetrical or skewed”. Croxton & Cowden Objectives of Skewness:- 1. To find out the nature and degree of concentration of values. Whether is higher of lower, it helps in decision making on the basis of distribution of wages, income etc. 2. To check as to what extent the empirical relations between mean, Median and Mode hold good. This empirical relationship is based on the moderately skewed distribution 3. To verify, if the distribution is normal many statistical measures are based on the assumption of normal distribution. Distinction between Dispersion and Skewness Dispersion Skewness 1. It is concerned with the amount of dispersion. 1. Skewness tells us about the directions of the variation or the departure from symmetry. 2. It give scattered ness of the observations. 2. It indicates to what extent and in what direction 3. Measure of dispersion does not depend on the the distribution differs from symmetry. measure of skewness. 3. Some measure of skewness depend on a 4. It judges the truthfulness of the measures of measures of dispersion. central Tendency. 4. It judges the differences between the measures 5. Measures of dispersion are the averages of of central Tendency second order 5. Measures of Skewness are not averages. TTESTS OF SKEWNESS: When a distribution in asymmetrical or, if skewness is present in a distribution, the following conditions are satisfied. 1. The values of Mean, Median and Mode do not coincide 2. When the values are plotted on a graph paper, they do not yield a normal or bell shaped form. i.e two equal halves are not found, when divided vertically through the centre of the curve. 3. Quartiles are not equidistant from the median i.e (Q3 –Md) is not equivalent to (Md – Q1) 4. The sum of the positive deviations from the median is not equal to the sum of the negative deviation. 5. Frequencies on either side of the mode are not equal. Measures of Skewness: 110
3. 3. There are four measures of skewness each consisting of absolute and relative measures. The relative measure is called the coefficient of skewness. The signs of the skewness indicate whether the distribution is positively skewed or negatively skewed. The four measures are as follows. 1. Karl Pearson’s co-efficient of Skewness 2. Bowley’s coefficient of skewness 3. Kelly’s measures Not included in the syllabus 4. Moments measures Karl Pearson’s coefficient of Skewness Karl Pearson gives the following formula to find out skewness. I. Absolute measure, sk=Mean – Mode If mode is not determined then Absolute measure sk =3(Mean – Median) II. Relative Measure Sk = 0 - Z (pearson’s first measure) σ If mode is ill – defined for a series, then alternative formula. Calculation of Karl Pearson’s coefficient of skewness ILLUSTRATION = 01 Find the coefficient of skewness from the data given below Size: 30 40 50 60 70 80 90 100 Frequency: 7 10 14 35 102 136 43 8 Solution: A=70, C=10, dx=(x – A)/C x f Dx fdx fd2x K.P. Coeff. Of Sk = 0 - Z 30 7 -4 -28 112 σ 40 -30 Maximum frequency is 136 hence the value of mode is80 10 -3 90 50 -28 14 -2 56 0 = A + Σfdx xC 60 -35 N 35 -1 35 70 0 = 70 + (125/355)x10 = 73.52 102 0 0 80 13 136 1 136 90 6 43 2 172 K.P coeff of Sk = 73.52 – 80 = 0.487 10 86 13.3 08 3 72 0 24 12 355 673 5 S.D =√Σfd2x -(Σfdx)2 x C N (N) = √ 673 - (125)2 x 10 355 (355) =√1.895 – (0.352)2 x 10 = √1.895 – 0.1239 x 10 = √1.7711 = 1.33 x 10 = 13.3 ILLUSTRATION = 02 Calculate Karl Pearsons’s coefficient of Skewness for the following data. Wages in Rs 15 20 40 45 55 250 300 350 500 600 0 0 0 0 0 No. of workers 05 15 10 20 12 8 20 5 3 2 111
4. 4. Solution K.P Coeff. Of Sk = 3(0 - M ) σ A= 350, C= 50 Wages By inspection, Mode is ill – defined since there are two f dx fdx fd2x cf In Rs. x maximum frequencies. 5 150 5 -4 -20 80 20 M = the size of (N+1th)/2 items value 200 15 -3 -45 135 30 = (100 + 1)/2 = 101/2 = 50.5 250 10 -2 -20 40 50 M = 350 300 20 -1 -20 20 62 0 = A + Σfdx xC 350 12 0 0 0 70 N 400 8 1 8 8 90 = 350 + (-20/100)x50 = 350 – (1000/100) = 350 –10 450 20 2 40 80 95 = 340 500 5 3 15 45 98 550 3 4 12 48 10 600 2 5 10 50 0 10 Σfdx -20 506 0 K.P coeff of Sk = 3(0 - M) S.D =√Σfd2x -(Σfdx)2 x C σ N (N) = 3(340 – 350) = √ 506 - (-20)2 x 50 112 100 (100) = (3 x –10)/112 =√5.06 – (−0.2)2 x 50 = −30/112 = −0.2678 = √5.06 – 0.04 x 50 = √5.02 x 50 = 2.24 x 50 = 112.00 ILLUSTRATION = 03 Calculate the Karl Pearson’s coefficient of Skewness from the following data Variable: 0–5 5 –10 10 –15 15 –20 20 –25 25 –30 30 –35 35 -40 Frequency 2 5 7 13 21 16 8 3 Solution K.P coeff of Sk = 0 - Z σ A = 17.5, C = 5, dx = (x – A)/C x F Midx dx fdx Fd2x Z = L1 + f1 – f0 (L2 – L1) 0–5 2 2.5 -3 -6 18 2f1 – f0 – f2 5 – 10 5 7.5 -2 -10 20 = 20 + 21 – 13 (25 – 20) 10 –15 7 12.5 -1 -7 7 2x21 –13 –16 15 –20 13 17.5 0 0 0 = 20 + 8 x5 = 20 +(40/13) = 23.07 20 –25 21 22.5 1 21 21 42 – 29 25 –30 16 27.5 2 32 64 0 = A + Σfdx xC 30 –35 8 32.5 3 24 72 N 35 –40 3 37.5 4 12 48 = 17.5 + (66/75)x5 = 17.5 +4.4 = 21.9 75 66 250 K.P coeff of Sk = 0 - Z S.D =√Σfd2x -(Σfdx)2 x C σ N (N) = (21.9 – 23.07)/7.993 = √ 250 - (66)2 x 5 = (−1.17/7.993) = -0.1463 75 (75) =√3.33 – (0.88)2 x 5 = √3.33 – 0.7744 x 5 112
5. 5. = √2.5556 x 5 = 1.5986 x5 = 7.993 ILLUSTRATION = 04 From the following data of the weekly wages of workers employed in a certain factory. Construct a frequency table with classes 40 – 49, 50 –59 and soon. Also calculate the Karl Pearson’s coefficient of skewness and co efficient of variable. 11 7 42 74 40 60 82 41 61 83 63 76 5 5 10 7 53 76 84 50 67 65 78 56 95 78 0 7 10 5 10 68 69 80 79 79 54 73 81 79 4 9 0 6 66 49 77 90 84 76 42 64 70 80 72 9 10 7 72 50 79 52 96 51 86 94 71 73 3 8 Solution (K.U.BBM) Table showing the distribution of weekly wages of workers employed in a certain factory. Calculation of coefficient of Skewness. A= 74.5, C = 10, dx = (x – A)/C Weekly 0 = A + Σfdx xC d Wages in Tally marks f Mid x fdx fd2x N x x = 74.5 + (-9/60)x10 1111 5 = 74.5 – (90/60) 40 – 49 1111 111 8 44.5 -3 -15 45 = 74.5 – 1.5 50 – 59 1111 1111 1 54.5 -2 -16 32 = 73 60 – 69 1111 1111 1111 1111 0 64.5 -1 -10 10 K.P coeff of Sk = 0 - Z 70 – 79 1111 111 2 74.5 0 0 0 σ 80 –89 1111 0 84.5 1 1 8 = (73.0-74.05)/16.41 90 – 99 1111 8 94.5 2 2 16 = (1.05/16.41) = 0.063 100 – 109 1 4 104.5 3 3 36 110 – 119 4 114.5 4 4 16 1 6 -9 -9 163 0 S.D =√Σfd2x -(Σfdx)2 x C Z = L1 + f1 – f0 (L2 – L1) N (N) 2f1 – f0 – f2 = √ 163 - (-9)2 x 10 L1= 70 –0.5 = 69.5, L2= 79 + 0.5 = 79.5 60 (60) = 69.5 + 20 – 10 (79.5 –69.5) =√2.717 – (-0.15)2 x 10 2x20 –10 –8 = √2.717 – 0.0225 x 10 = 69.5 + 10 x10 =69.5+(100/22) = 74.05 = √2.6945 = 1.641 x10 = 16.41 40 – 18 ILLUSTRATION = 05 From the following data calculate Karl Pearson’s Coefficient of Skewness and coefficient of variation. Production in tonnes 90 80 60 50 30 20 Less than 1000 700 400 100 0 0 0 0 0 0 No. of Firms 95 82 62 53 32 19 1000 750 400 100 0 0 0 0 0 0 (K.U.BBM) Solution Convert the given cumulative frequency distribution into an ordinary Table. 113
6. 6. Mid d Since there are four maximum frequencies, by x f fdx fd2x cf x x inspection, the value of mode is ill – defined. So 0 –100 100 50 -4 -400 1600 100 alternative formula for Skewness calculation is as 100-200 90 150 -3 -270 810 190 under 200-300 130 250 -2 -260 520 320 300-400 80 350 -1 -80 80 400 M = L1 + L2 – L1 (m – c) 400-500 130 450 0 0 0 530 f 500-600 90 550 1 90 90 620 Where m = N/2 =1000/2 =500 600-700 130 650 2 260 520 750 = 400 + 500 – 400(500- 400) 700-800 70 750 3 210 630 820 130 800-900 130 850 4 520 2080 950 = 400 +(100/130)x 100 900-1000 50 950 5 250 1250 1000 = 400 + (10000/130) = 400 + 76.92 100 = 476.92 320 7580 0 S.D =√Σfd2x -(Σfdx)2 x C 0 = A + Σfdx xC N (N) N = √ 7580 - (320)2 x 100 = 450 + (320/1000)x100 1000 (1000) = 450 + 32 =√7.58 – (0.32)2 x 100 = 482 = √7.58 – 0.1024 x 100 = √7.4776 = 2.734 x 100 = 273.4 K.P coeff of Sk = 3(0 - M) σ = 3(482-476.92)/273.4 = 3 x 5.08/273.4 = 25.4/273.4 = 0.092 CV = (2734/482)x100 = 56.72% ILLUSTRATION = 06 From the following data, calculate the value of Karl Pearson’s coefficient of Skewness and coefficient of variation. Marks above 4 6 05 15 25 35 55 75 85 5 5 No. of Students 13 12 11 8 2 140 45 10 03 3 5 5 5 0 Solution : Convert the given cumulative frequency distribution into an ordinary table. A=50, C=10, dx = (x – A)/c x F Midx dx fdx fd2x Z = L1 + f1 – f0 (L2 – L1) 5 – 15 7 10 -4 -28 112 2f1 – f0 – f2 15 –25 8 20 -3 -24 72 = 45 + 40 – 30 (55 –45) 25 –35 10 30 -2 -20 40 2x40 –30 –25 35 –45 30 40 -1 -30 30 = 45 + 10 x10 = 45+(100/25) = 49 45 –55 40 50 0 0 0 80 – 55 55 –65 25 60 1 25 25 0 = A + Σfdx xC 65 –75 10 70 2 20 40 N 75 –85 07 80 3 21 63 = 50+ (-24/140)x10 = 50 – (240/140) 85 –95 03 90 4 12 48 = 50 – 1.71 = 48.29 14 -24 430 0 K.P coeff of Sk = 0 - Z S.D =√Σfd2x -(Σfdx)2 x C σ N (N) = (48.29 – 49)/17.44 = √ 430 - (-24)2 x 10 114
7. 7. = -0.0040710 140 (140) =√3.071– (-0.171)2 x 10 = √3.071– 0.029 x 10 = √3.042 x 10 = 1.744 x10 = 17.44 ILLUSTRATION = 07 Calculate Karl Pearson’s coefficient of Skewness from the following data. Marks above 4 0 10 20 30 50 60 70 80 0 No. of Students 15 14 10 8 80 70 30 14 0 0 0 0 0 Solution :- Convert the cumulative frequency distribution into ordinary frequency distribution A=45, C =10, dx=(x – A)/C X f cf Midx dx fdx fd2x Note : Here frequencies are irregular, so mode is ill – 10 defined and we use alternative formula for skewness. 50 Sk = 0 - Z 0 – 10 10 70 5 -4 -40 160 σ 10 - 20 40 70 15 -3 -120 360 M = L1 + L2 – L1 (m – c) 20 –30 20 80 25 -2 -40 80 f 30 –40 0 12 35 -1 0 0 Where m = N/2 =150/2 =75 40 –50 10 0 45 0 0 0 = 40 + 50 – 40(75 -70) 50 –60 40 13 55 1 40 40 10 60 –70 16 6 65 2 32 64 = 40 +(10/10)x 5 70 –80 14 15 75 3 42 126 = 45 80 –90 0 0 85 4 0 0 15 0 15 -86 830 0 0 = A + Σfdx xC S.D =√Σfd2x -(Σfdx)2 x C N N (N) = 45+ (-86/150)x10 = 45 – (860/150) = 39.27 = √ 830 - (-86)2 x 10 150 (150) K.P coeff of Sk = 3(0 - M) =√5.533 – (0.57)2 x 10 σ = √5.2081 x 10 = 3(39.27 -45)/22.82 = 17.19/22.82 = 2.282 x 10 = 22.82 = 0.7532 Coefficient of Variabtion = (σ/0)x100 = (22.82/39.27)x100 = 58.11% ILLUSTRATION = 08 Find standard deviation and Karl Pearson’s coefficient of skewness. Weight in lbs 70 –79.9 80 –89.9 90 –99.9 100 –109.9 110-119.9 120-129.9 No. of persons 12 18 35 42 50 45 Solution:- Convert the given inclusive series into exclusive series. A = 94.5, C= 10, dx = (x – A)/C 115
8. 8. X f Midx dx fdx fd2x 0 = A + Σfdx xC 69.95 –79.95 12 74.95 -2 -24 48 N 79.95 –89.95 18 84.95 -1 -18 18 = 94.95+ (235/202)x10 = 94.95 +11.63 89.95 –99.95 35 94.95 0 0 0 = 106.58 99.95 –109.95 42 104.95 1 42 42 109.95-119.95 50 114.95 2 100 200 Maximum frequency is 50, so modal class is 119.95 –129.95 45 124.95 3 135 405 109.95 – 119.95 202 235 713 S.D =√Σfd2x -(Σfdx)2 x C Z = L1 + f1 – f0 (L2 – L1) N (N) 2f1 – f0 – f2 = √ 713 - (235)2 x 10 = 109.95+ 50 – 42 (119.95– 109395) 202 (220) 2x50 –42 –45 =√3.5297– (1.163)2 x 10 = 109.95+ 8 x10 = 109.95+6.15 = 116.1 = √3.5297 –1.352 x 10 100 -87 = 1.475 x 10 = 14.75 K.P coeff of Sk = 0 -Z σ Coefficient of Variabtion = (σ/0)x100 = (106.58 – 116.1 )/14.75 = (14.75/106.58)x100 = 13.83% = - 9.52/14.75 = = -0.645 ILLUSTRATION = 9 In a certain distribution, the following results were established: Mean = 45, Median = 48, Sk = -0.4 Find the standard deviation. Solution Skp = 3(Mean – Median) σ - 0.4 = 3(45 – 48) = -0.4 = (3 x –3) σ σ σ = -9/-0.4 = 22.5 ILLUSTRATION = 10 Compute the coefficient of Skewness form the following : Mean = 59.50, Median = 55.70, Variance = 110 Solution σ = √Variance = σ = √110 = 10.488 Skp = 3(0 - M) = 3(59.50-55.70) = 1.08695 σ 10.488 ILLUSTRATION = 11 Calculate coefficient of skewness, if Σfx = 350, n = 10, Median =38 and Variance = 49 Solution σ = √Variance = σ = √49 = 7 S.D 0 = (Σfx)/n =350/10 = 35 Skp = 3(0 -M) = 3(35 –38) = (3x –3)/7 = -1.2857 σ 7 ILLUSTRATION = 12 For a moderately skewed data, the arithmetic mean is 200, the coefficient of variation is and Karl Pearson’s of co efficient of skewness is 0.3. Find the mode and the median. Solution: - Standard deviation can be determined by using the coefficient of variation formula C.V=(σ/0)x100, 8 = (σ/200)x100 = S.D.= 16 116
9. 9. Mode can be determined by using the formula of coefficient of skewness Skp = (0 - Z)/σ, 0.3 = (200-Z)/16 = Mode = 195.2 Median can be determined by using the formula Mode = 3M - 20 195.2= 3M – 2x200 195.2 = 3 – 400 M = (595.2/3) M = 198.4 ILLUSTRATION = 13 The coefficient of skewness for a certain distribution is 0.35. The median and mode are 27.4 and 25.3 respectively, calculate the coefficient of variation. Solution: C.V = (σ x 0)x100 Hear standard deviation and mean values are not given To find out mean value use the formula Z = 3M - 20 25.3 = 3 X27.4 - 20 25.3 = 82.2 - 20 = -20= 82.2 – 25.3, 0=(56.9/2)=28.45 To find out S.D use Skewness formula Skp = (0 -Z)/σ = 0.35 = (28.45 – 25.3)/σ 0.35σ = 3.15, σ = (3.15/0.35) = 9 ∴C.V = (9/28.45)/100 = 31.63% ILLUSTRATION = 14 The mean, mode and coefficient of skewness for a certain distribution are 34.8, 30.6 and 0.7 respectively calculate coefficient of variation. Solution CV= (σ/0)x100 To find out S.D use the formula = (5.57/34.5)x100 S.K = (0 - Z)/σ = 0.7 = (34.5 – 30.6)/ σ = 16.14% 0.7 = (3.9/σ) = σ = (3.9/0.7) = 5.57 ILLUSTRATION = 15 You are given, Mean = 50, CV=40%, Skp = -0.4 Find out a. Standard Deviation b. Mode Solution To find out standard deviation use CV formula CV = (σ/0)x100, 40 = (σ/50)x100, σ = (40 x 50)/100 = 20 To find out mode use Skp formula Skp = (0 - Z)/σ, -0.4 = (50 –Z)/20 = Mode = 58 -Z= -50 –8.0 Z = (-58/-1) = Z = 58 BOWLEY’S COEFFICIENT OF SKEWNESS Prof. Bowley have suggested a formula based on relative position of quartiles. In a symmetrical, the quartiles are equidistant from the value of the mean i.e. Median – Q1 = Q3 – Median . But in a skewed distribution, the quartiles will not be equidistant from the median. Hence Bowley has suggested the following formula. B.coeff. of sk= sum of two Quartiles – 2 Median Difference of two quartiles B.Coeff. of sk = Q3 + Q1 - 2M 117
10. 10. Q3 –Q1 Calculation of Bowley’s coefficient of skewness and quartile deviation ILLUSTRATION = 01 From the data given below, calculate Bowley’s coefficient of skewness and coefficient of quartile deviation. Wages in Rs 125 130 135 140 145 150 155 160 165 170 No. of Workers 10 12 8 5 15 13 7 14 6 10 Solution X f cf Median = The size of (N+1)/2, (100+1)/2 = 50.5 125 10 10 M = 150 130 12 22 Q1 = The size of (N+1)th/4 item’s values 135 8 30 25.25 = (100+1)/4 = 25.25 140 5 35 Q1 = 135 145 15 50 Q3 = The size of 3(N+1)th/4 items value 150 13 63 50.5 = 3(100+1)/4 = (3x101)/4 = 75.75 155 7 70 Q3 = 160 160 14 84 75.75 1. B.Coeff.sk = (Q +Q - 2M)/Q – Q 3 1 3 1 165 6 90 = (160+135-2x150)/(160-135) = (295 –300)/25 170 10 100 = 5/25 = 0.2 N 100 2. Coefficient of Q.D =(Q3 – Q1)/(Q3 +Q1) = (160 –135)/(160+135) = 25/295 =0.08 ILLUSTRATION = 02 From the following compute Bowley’s coefficient of Skewness and Quartile Deviation. Weekly 40 –60 80 –120 120-160 160-200 200-240 240-280 280-320 320-360 wages No. of 8 12 40 30 20 30 40 20 workers Solution = 02 x F cf Median = L1+ L2 – L1(m – c) Wheere m = N/2= 200/2= 100 40 –80 8 8 f 80 –120 12 20 50 =200+240-200(100-90) 120 –160 40 60 20 160 –200 30 90 100 = 200 +(40/20)x10 = 200+20 = 220 200 –240 20 110 Q1 = L1+L2 – L1(q1 –c)Where q1 = N/4 = 200/4 = 50 240 –280 30 140 150 f 280 –320 40 180 = 120+ 160-120(50-20) 320 -360 20 200 40 200 = 120+(40/40)x30 = 150 B.Coeff of sk = Q3 + Q1 – 2M Q3= L1+ L2 –L1 (q3 – c) Q3 – Q1 f = (290 + 150 –2x220)/290-150 Where q3 =3N/4 = (3x200)/4=150 = (440 –440)/140 = 0 = 280+320 -280(150-140) 40 = 280 +(40/40)x10 = 290 ILLUSTRATION = 03 Calculate Bowley’s coefficient skewness from the following data. Mid value 75 100 125 150 175 200 225 250 Frequency 35 40 48 100 125 80 50 22 (K.U. BBM) 118
11. 11. Solution The class interval is 25. To find out the classes deduct from Mid value and add 12.5 with the mid value. X F Cf Q1 = L1+L2 – L1(q1 –c)Where q1 = N/4 = 500/4 = 125 62.5 –87.5 35 35 = 137.5+ 162.5- 137.5(125-123) 87.5 –112.5 40 75 100 112.5 –137.5 48 123 12 = 137.5+(25/100)x2 = 138 135.5 –162.5 100 223 5 162.5 –187.5 125 348 Median = L1+ L2 – L1(m – c) Wheere m = N/2= 200/2= 100 187.5 –212.5 80 428 25 f 212.5 –237.5 50 478 0 =162.5+187.5 –162.5(250 –223) 237.5 –267.5 22 500 37 125 5 = 167.9 500 Q3= L1+ L2 –L1 (q3 – c) B.Coeff of sk = Q3 + Q1 – 2M f Q3 – Q1 Where q3 =3N/4 = (3x500)/4=375 = (195.94+138 –2x167.9)/195.94 –138 = 187.5+212.5 -187.5(375 –348) =-1.86/57.94 = -0.032 80 = 195.94 ILLUSTRATION = 04 From the following data of weekly wages of workers employed in a factory. From a frequency distribution table using inclusive method and from the first class as 10 – 19 with 10 as the magnitude of class interval. Also calculate coefficient of quartile deviation and Bowley’s coefficient of skewness. 32 61 52 56 22 49 97 35 30 20 95 67 42 20 31 64 20 10 60 62 27 53 31 19 54 25 43 47 21 35 43 75 45 22 36 13 46 23 11 51 15 39 50 42 77 73 81 40 55 40 Solution (K.U.BBM) Preparation of frequency Table in inclusive form and computation of Bowley’s coefficient of skewness. x Tally Marks f cf Median = L1+ L2 – L1(m – c) Where m = N/2= 50/2= 25 1111 5 5 f 10 -19 12. 1111 1111 9 14 5 L1 = 40 –0.5 = 39.5 20 – 29 1111 111 8 22 L2 = 49 +0.5 = 49.5 30 –39 1111 1111 1 32 25 =39.5+49.5 –39.5(25 – 22) 40 –49 1111 11 0 39 37. 10 50 –59 1111 7 44 5 = 39.5 +(10/10)x 3 = 42.5 60 –69 111 5 47 Q1 = L1+L2 – L1(q1 –c)Where q1 = N/4 = 50/4 = 12.5 70 – 79 1 3 48 f 80 –89 11 1 50 L1 = 20 –0.5=19.5, L2= 29+0.5= 29.5 90 –99 2 = 19.5+ 29.5 – 19.5(12.5 - 5) 5 9 0 = 19.5+(10/9)x7.5 = 19.5+8.33=27.83 Q3= L1+ L2 –L1 (q3 – c) B.Coeff of sk = Q3 + Q1 – 2M f Q3 – Q1 Where q3 =3N/4 = (3x50)/4=37.5 = (57.35+27.83 –2x42.5)/57.35 –27.83 119
12. 12. L1 = 50 – 0.5=49.5, L2 = 59+0.5= 59.5 =(85.18 –85.0)/29.52 =0.18/29.52 = 49.5+59.5 -49.5(37.5 –32) = 0.00609 7 = 49.5 + (10/7)x5.5 Q.D = (Q3 – Q1)/2 = (57.35 –27.83)/2 = 49.5+7.85 = 57.35 = 14.76 ILLUSTRATION = 05 Calculate coefficient of Quartile Deviation and Coefficient of skewness from the data given below. Income in Rs. 60 Below 400 500 700 800 900 1000 1100 1200 0 No. of workers 11 14 60 194 262 324 372 394 400 8 (K.U.BBM) Solution: Convert the given cumulative frequency distribution into an ordinary frequency Table, X F cf Median = L1+ L2 – L1(m – c) Where m = N/2= 400/2= 200 14 f 60 10 =700 +800 –700(200 – 194) 11 0 68 8 = 708.82 300 –400 14 19 20 Q1 = L1+L2 – L1(q1 –c)Where q1 = N/4 = 400/4 = 100 400 –500 46 4 0 f 500 –600 58 26 = 500+ 600 – 500(100 –60) 600 –700 76 30 2 58 700 –800 68 0 32 Q1= 568.96 800 –900 62 4 900 –1000 48 37 1000-1100 22 2 1100 –1200 06 39 4 40 0 N= 400 Q3= L1+ L2 –L1 (q3 – c) B.Coeff of sk = Q3 + Q1 – 2M f Q3 – Q1 Where q3 =3N/4 = (3x400)/4=300 = (861.29+568.96 –2x708.82)/861.29-568.96 = 800+900 – 800(300 –262) = (1430.25 –1417.64)/292.33 62 = 0.043 = 861.29 Q.D = (Q3 – Q1)/2 = (861.29 –568.96)/2 = 292.33/2 = 146.165 ILLUSTRATION= 6 From the following data calculate coefficient of Quartile Deviation and hence coefficient of Quartile deviation and hence coefficient of skewness. Income in Rs. 20 40 80 Above 0 600 1000 1200 1400 1600 1800 0 0 0 No. of Families 49 47 36 500 435 240 150 85 45 20 0 0 0 (K.U. BBM) Solution Convert the given cumulative frequency distribution into an ordinary frequency distribution table. 120
13. 13. Income Median = L1+ L2 – L1(m – c) Where m = N/2= 500/2= 250 in Rs f cf f x =800+1000 –800(250 – 140) 10 120 30 = 983.33 65 Q1 = L1+L2 – L1(q1 –c)Where q1 = N/4 = 500/4 = 125 12 10 14 f 0 –200 5 20 0 = 600+ 800 – 600(125 –65) 200 –400 25 35 26 75 400 –600 0 75 0 = 600 +(200/75)x60 600 –800 12 35 =760 800 –1000 37 0 0 1000 –1200 5 90 41 1200 –1400 65 5 1400 –1600 40 45 1600 –1800 25 5 1800 –2000 20 48 0 50 0 50 0 Q3= L1+ L2 –L1 (q3 – c) B.Coeff of sk = Q3 + Q1 – 2M f Q3 – Q1 Where q3 =3N/4 = (3x500)/4=375 = (1276.92+760 –2x983.33)/1276.92 –760 = 1200+1400 – 1200(375 –350) = 0.1359 65 = 1200+(200/65)x25 = 1276.92 Q.D = (Q3 + Q1 –2M)/Q3 – Q1 = (1276.92 –760)/(1276.92 +760) = 516.92/2036.92 = 0.253 ILLUSTRATION = 07 In a frequency distribution the coefficient of skewness based on quartiles is 0.6. If the sum of the upper and lower quartiles is 100 and median is 38. find the value of the upper quartile Solution Given data :- coeff. Sk= 0.6, Q3 + Q1= 100, M = 38 SkB = (Q3 +Q1 – 2M)/Q3 – Q1 = 0.6 = (100 –2x38)/Q3 – Q1 Let x represents the difference between Q3 & Q1 ∴0.6 = (100 –76)/x = x= (24/0.4)=40 Using equation Q3 Value will be :- Q3 + Q1 = 100 Q3 – Q1 = 40 by adding 2Q3 = 140 ∴Q3 = (140/2) Q3 = 70 ∴Q1 = 100 –70 = 30 ILLUSTRATION = 08 Find the coefficient of skewness, if difference between two quartiles is 8, and their sum is 22. Median is 10.5. B.coeffi of sk = (Q3 +Q1 – 2M)/Q3 –Q1 = (22 –2x10.5)/8 = 22 – 21/8 = 0.125 To find out Q1 & Q3 use simultaneous equation 121
14. 14. Q3 + Q1 = 22 Q3 – Q1 = 8 by adding 2Q3 = 30 ∴Q3 = 30/2 =15, Q1 = 22 – 15 =7 ILLUSTRATION = 09 For a distribution Bowley’s coefficient of Skewness is –0.36 Q1, is 8.6 and Median is 123. Find the value of upper quartile. (K.U. BBM) Solution SkB = (Q3 + Q1 = 2M)/Q3 –Q1) = -0.36=(Q3 – 8.6 – 2x12.3)/Q3 – 8.6 = -0.36 = (Q3 +8.6 –24.6)/Q3 – 8.6 = (Q3 + 8.6 – 24.6)/Q3 – 8.6 = -0.36(Q3 – 8.6) = Q3 + 8.6 – 24.6 =-0.36Q3 + 3.096= Q3 +8.6 –24.6 -1.36Q3 = +8.6 – 24.6 –3.096 -1.36Q3 = 8.6 – 27.696 -1.36Q3 = -19.096 ∴= -19.096/-1.36 = Q3 = 14.04 ILLUSTRATION = 10 In a distribution the coefficient of skewness if –0.33 and the difference between two quartiles is 15 and their sum is 35. Find the value of Median. Solution B.coeff.of sk = (Q3+Q1 – 2M)/Q3 – Q1 -0.33 = (35 –2M)/15 = 35 – 2M=-4.95 -2M = -4.95 –35 M = -39.95/ -2 = 19.975 THEORETICAL QUESTIONS(5, 10 & 15 Marks) 1. What do you understand by skewness? What are the various methods of measuring skewness? 2. Explain the concept of skewness Distinguish between positive and negative skewness. 3. What is Skewness? What are the tests of skewness? 4. When is skewness present in a series? 5. Distinguish between dispersion and skewness. PRACTICAL PROBLEMS 1. Calculate from the following data the measure of skewness bases on Mean, Median and standard Deviation. Values 100-200 200-300 300-400 400-500 500-600 600-700 700-800 800-900 Frequency 45 88 146 206 79 52 30 14 [Ans0 = 469.39, SD = 158.5, M =424.76, sk = 0.845] 2. Calculate coefficient of skewness based on Mean, Median and standard deviation from the following data. Marks Below 10 20 30 40 50 60 70 80 No. of Students 14 19 6 18 40 88 176 200 4 4 [Ans0 = 41.7, M = 42.14, SD = 15.43, Sk = -0.086] 3. Calculate Karl Pearson’s coefficient of skewness from the following data. Marks Above 5 15 25 35 45 55 65 75 85 No. of students 10 120 96 85 72 58 32 12 0 5 [ Ans0 = 48.33, M = 53.57, SD = 22.15, Sk = -0.71] 4. Calculate Karl Pearson’s coefficient of skewness from the following data. Midvalue 75 85 95 105 115 125 135 145 Frequency 12 18 35 42 50 45 20 80 [Ans 0 = 110.43, Z =116.15, SD = 17.26, sk= -0.331] 122
15. 15. 5. Calculate the coefficient of variation & Karl Pearson’s coefficient of skewness from the following data. Vales 1 –10 11 –20 21 –30 31 –40 41 –50 51 –60 61 –70 71 –80 Frequency 5 10 15 25 26 19 20 5 [ Ans 0 = 42.52, SD = 17.69, Z = 41.25, Sk = 0.07] Problems On Bowley’s Method 6. Calculate coefficient of skewness bases on quartiles from the following frequency distribution. Mid Value 5 15 25 35 45 55 65 75 Frequency 10 40 20 0 10 40 16 14 [Ans Q1 = 16.875, Q2 = 42.5, Q3 = 58.125. SkB = -0.24] 7. Calculate Bowley’s coefficient of skewness and quartile deviation from the following data. Variable 10 –19 20 –29 30-39 40-49 50-59 60-69 70-79 80-89 Frequency 5 9 14 20 25 15 8 4 [Answer: Q1 = 37.4 Q2 = 50.3, Q3=60.8, Sk = -0.103] 8. Calculate Bowley’s Coefficient of Skewness from the following distribution. Age in Years 2 Below 10 15 25 30 35 40 50 0 No. of workers 8 17 22 26 15 43 251 275 2 7 7 7 [Q1 = 18.3013, Q3 = 27.925, M = 22.92, Sk = 0.0399] 9. Calculate Bowley’s coefficient of skewness from the following data. Marks 5 Above 5 15 25 35 45 65 75 85 95 5 Students 12 11 10 6 123 87 43 23 7 2 1 2 1 7 [Ans: Q3 = 71.125, Q1 = 40.71, M = 57.29, Sk = -0.0902] 10. From the following data, calculate Bowley’s coefficient of skewness and quartile deviation. Wages in No. of Workers Rs. x f Upto 100 15 100 –109.9 20 110 –119.9 24 120 –129.9 18 130 –139.9 32 140 –149.9 26 150 –159.9 28 160 –169.9 22 170 –179.9 15 180 & above 10 N= 210 123