™“Statistics may be rightly called the science of averages and their estimates.” – A.L.BOWLEY & A.L. BODDINGTON • 15.1 Introduction We know that statistics deals with data collected for specific purposes. We can make decisions about the data by analysing and interpreting it. In earlier classes, we have studied methods of representing data graphically and in tabular form. This representation reveals certain salient features or characteristics of the data. We have also studied the methods of finding a representative value for the given data. This value is called the measure of central tendency. Recall mean (arithmetic mean), median and mode are three measures of central tendency. A measure of central tendency gives us a rough idea where data points are centred. But, in order to make better interpretation from the data, we should also have an idea how the data are scattered or how much they are bunched around a measure of central tendency. Consider now the runs scored by two batsmen in their last ten matches as follows: Batsman A : 30, 91, 0, 64, 42, 80, 30, 5, 117, 71 Batsman B : 53, 46, 48, 50, 53, 53, 58, 60, 57, 52 Clearly, the mean and median of the data are Batsman A Batsman B Mean 53 53 Median 53 53 Recall that, we calculate the mean of a data (denoted by x ) by dividing the sum of the observations by the number of observations, i.e., x = 1 ∑ n xi n i =1 Also, the median is obtained by first arranging the data in ascending or descending order and applying the following rule. 1 th⎛n +⎞If the number of observations is odd, then the median is ⎜⎟observation.⎝2 ⎠⎛⎞th nIf the number of observations is even, then median is the mean of ⎜⎟ and2⎝⎠⎛n ⎞th ⎜+1⎟observations.⎝2 ⎠We find that the mean and median of the runs scored by both the batsmen A and B are same i.e., 53. Can we say that the performance of two players is same? Clearly No, because the variability in the scores of batsman A is from 0 (minimum) to 117 (maximum). Whereas, the range of the runs scored by batsman B is from 46 to 60. Let us now plot the above scores as dots on a number line. We find the following diagrams: For batsman A For batsman B Fig 15.1 Fig 15.2 We can see that the dots corresponding to batsman B are close to each other and are clustering around the measure of central tendency (mean and median), while those corresponding to batsman A are scattered or more spread out. Thus, the measures of central tendency are not sufficient to give complete information about a given data. Variability is another factor which is required to be studied under statistics. Like ‘measures of central tendency’ we want to have a single number to describe variability. This single number is called a ‘measure of dispersion’. In this Chapter, we shall learn some of the important measures of dispersion and their methods of calculation for ungrouped and grouped data. 15.2 Measures of Dispersion The dispersion or scatter in a data is measured on the basis of the observations and the types of the measure of central tendency, used there. There are following measures of dispersion: (i) Range, (ii) Quartile deviation, (iii) Mean deviation, (iv) Standard deviation. In this Chapter, we shall study all of these measures of dispersion except the quartile deviation. 15.3 Range Recall that, in the example of runs scored by two batsmen A and B, we had some idea of variability in the scores on the basis of minimum and maximum runs in each series. To obtain a single number for this, we find the difference of maximum and minimum values of each series. This difference is called the ‘Range’ of the data. In case of batsman A, Range = 117 – 0 = 117 and for batsman B, Range = 60 – 46 = 14. Clearly, Range of A > Range of B. Therefore, the scores are scattered or dispersed in case of A while for B these are close to each other. Thus, Range of a series = Maximum value – Minimum value. The range of data gives us a rough idea of variability or scatter but does not tell about the dispersion of the data from a measure of central tendency. For this purpose, we need some other measure of variability. Clearly, such measure must depend upon the difference (or deviation) of the values from the central tendency. The important measures of dispersion, which depend upon the deviations of the observations from a central tendency are mean deviation and standard deviation. Let us discuss them in detail. 15.4 Mean Deviation Recall that the deviation of an observation x from a fixed value ‘a’ is the difference x – a. In order to find the dispersion of values of x from a central value ‘a’ , we find the deviations about a. An absolute measure of dispersion is the mean of these deviations. To find the mean, we must obtain the sum of the deviations. But, we know that a measure of central tendency lies between the maximum and the minimum values of the set of observations. Therefore, some of the deviations will be negative and some positive. Thus, the sum of deviations may vanish. Moreover, the sum of the deviations from mean ( x ) is zero. Sum of deviations 0 = == 0Also Mean of deviations Number of observations n Thus, finding the mean of deviations about mean is not of any use for us, as far as the measure of dispersion is concerned. Remember that, in finding a suitable measure of dispersion, we require the distance of each value from a central tendency or a fixed number ‘a’. Recall, that the absolute value of the difference of two numbers gives the distance between the numbers when represented on a number line. Thus, to find the measure of dispersion from a fixed number ‘a’ we may take the mean of the absolute values of the deviations from the central value. This mean is called the ‘mean deviation’. Thus mean deviation about a central value ‘a’ is the mean of the absolute values of the deviations of the observations from ‘a’. The mean deviation from ‘a’ is denoted as M.D. (a). Therefore, Sum of absolute values of deviations from 'a' M.D.(a) = .Number of observations Remark Mean deviation may be obtained from any measure of central tendency. However, mean deviation from mean and median are commonly used in statistical studies. Let us now learn how to calculate mean deviation about mean and mean deviation about median for various types of data 15.4.1 Mean deviation for ungrouped data Let n observations be x1, x2, x3, ...., xn. The following steps are involved in the calculation of mean deviation about mean or median: Step 1 Calculate the measure of central tendency about which we are to find the mean deviation. Let it be ‘a’. Step 2 Find the deviation of each xi from a, i.e., x– a, x– a, x– a,. . . , x – a1 2 3 n Step 3 Find the absolute values of the deviations, i.e., drop the minus sign (–), if it is there, i.e., x1 − a , x2 − a , x3 − a ,...., xn − a Step 4 Find the mean of the absolute values of the deviations. This mean is the mean deviation about a, i.e., ∑ n xi − a i=1M.D.() =a n xi − xThus M.D. ( x ) =1 ∑ n , where x = Mean n i=1 xi − Mand M.D. (M) = 1 ∑ n , where M = Mediann i=1 $Note In this Chapter, we shall use the symbol M to denote median unless stated otherwise.Let us now illustrate the steps of the above method in following examples. Example 1 Find the mean deviation about the mean for the following data: 6, 7, 10, 12, 13, 4, 8, 12 Solution We proceed step-wise and get the following: Step 1 Mean of the given data is 6 ++ + + +++ 71012134812 72 x = ==9 88 Step 2 The deviations of the respective observations from the mean x, i.e., xi – x are 6 – 9, 7 – 9, 10 – 9, 12 – 9, 13 – 9, 4 – 9, 8 – 9, 12 – 9, or –3, –2, 1, 3, 4, –5, –1, 3 Step 3 The absolute values of the deviations, i.e., xi −x are 3, 2, 1, 3, 4, 5, 1, 3 Step 4 The required mean deviation about the mean is 8 8 3 +++++++ 2134513 22 = ==275 . 88 $Note Instead of carrying out the steps every time, we can carry on calculation, step-wise without referring to steps. Example 2 Find the mean deviation about the mean for the following data : 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5 Solution We have to first find the mean ( x ) of the given data 1 20 200x = 20 ∑xi = = 10 i=1 20 xi −xThe respective absolute values of the deviations from mean, i.e., are 2, 7, 8, 7, 6, 1, 7, 9, 10, 5, 2, 7, 8, 7, 6, 1, 7, 9, 10, 5 20 xi −x =124Therefore ∑ i=1 124 and M.D. ( x ) = = 6.220 Example 3 Find the mean deviation about the median for the following data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. Solution Here the number of observations is 11 which is odd. Arranging the data into ascending order, we have 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 ⎛11 +1⎞th Now Median = or 6th observation = 9⎜⎟⎝ 2 ⎠xi −M areThe absolute values of the respective deviations from the median, i.e., 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12 11 xi −M58 =Therefore ∑ i=1 11 1and ()= 1 ∑xi −MM.D. M =×58 =5.27 11 i=1 11 15.4.2 Mean deviation for grouped data We know that data can be grouped into two ways : (a)Discrete frequency distribution, (b)Continuous frequency distribution. Let us discuss the method of finding mean deviation for both types of the data. (a) Discrete frequency distribution Let the given data consist of n distinct values x1, x, ..., x occurring with frequencies f1, f, ..., f respectively. This data can be 2n2 nrepresented in the tabular form as given below, and is called discrete frequency distribution: x : x1 x2 x3 ... xn f : f ff ... f123n (i) Mean deviation about mean First of all we find the mean x of the given data by using the formula n xf∑ ii 1 n i=1x== xf∑ iin N,i=1∑ fi i=1 n where ∑ xifi denotes the sum of the products of observations xi with their respective i=1n frequencies fi and N=∑ fi is the sum of the frequencies. i=1 Then, we find the deviations of observations xifrom the mean xand take their absolute values, i.e., xi− xfor all i=1, 2,..., n. After this, find the mean of the absolute values of the deviations, which is the required mean deviation about the mean. Thus n ∑fi i=1 (ii) Mean deviation about median To find mean deviation about median, we find the median of the given discrete frequency distribution. For this the observations are arranged in ascending order. After this the cumulative frequencies are obtained. Then, we identify N the observation whose cumulative frequency is equal to or just greater than , where2 N is the sum of frequencies. This value of the observation lies in the middle of the data, therefore, it is the required median. After finding median, we obtain the mean of the absolute values of the deviations from median.Thus, 1 n M.D.(M) = fxi− M N ∑ i i=1 Example 4 Find mean deviation about the mean for the following data : xi 2 5 6 8 10 12 fi 28 10 7 8 5 Solution Let us make a Table 15.1 of the given data and append other columns after calculations. Table 15.1 xi fi fixi xxi − fi xxi − 2 2 4 5.5 11 5 8 40 2.5 20 6 10 60 1.5 15 8 7 56 0.5 3.5 10 8 80 2.5 20 12 5 60 4.5 22.5 40 300 92 66 6 N =∑ fi = 40 , ∑ fixi =300 , ∑ fi x− x =92i i=1 i=1 i=1 x = fx =× 300 = 7.5 Therefore 1 ∑ 6 ii 1 N 40 i=1 16 1 xi − xand M.D.() x =∑ fi =× 92 = 2.3N 40 i=1 Example 5 Find the mean deviation about the median for the following data: Solution The given observations are already in ascending order. Adding a row corresponding to cumulative frequencies to the given data, we get (Table 15.2). Table 15.2 xi 3 6 9 12 13 15 21 22 fi 3 4 5 2 4 5 4 3 c.f. 3 7 12 14 18 23 27 30 Now, N=30 which is even. Median is the mean of the 15th and 16th observations. Both of these observations lie in the cumulative frequency 18, for which the corresponding observation is 13. th th15 observation +16 observation 13 +13Therefore, Median M = == 13 22 Now, absolute values of the deviations from median, i.e., xi−M are shown in Table 15.3. Table 15.3 Mix− 10 7 4 1 0 2 8 9 fi 3 4 5 2 4 5 4 3 fi Mix− 30 28 20 2 0 10 32 27 88 We have ∑ fi = 30 and ∑ fi xi− M = 149 i=1 i=1 Therefore M. D.(M) = 18 fxi− MN ∑i i=1 1 = ×149 = 4.97.30 (b)Continuous frequency distribution A continuous frequency distribution is a series in which the data are classified into different class-intervals without gaps alongwith their respective frequencies. For example, marks obtained by 100 students are presented in a continuous frequency distribution as follows : (i)Mean deviation about mean While calculating the mean of a continuous frequency distribution, we had made the assumption that the frequency in each class is centred at its mid-point. Here also, we write the mid-point of each given class and proceed further as for a discrete frequency distribution to find the mean deviation. Let us take the following example. Example 6 Find the mean deviation about the mean for the following data. Solution We make the following Table 15.4 from the given data : Table 15.4 Marks obtained Number of students fi Mid-points xi fixi x xi− fi x xi− 10-20 2 15 30 30 60 20-30 3 25 75 20 60 30-40 8 35 280 10 80 40-50 14 45 630 0 0 50-60 8 55 440 10 80 60-70 3 65 195 20 60 70-80 2 75 150 30 60 40 1800 400 77 7 Here N =∑fi =40, ∑fii x=1800, ∑fixx =400i− i=1 i=1 i=1 1 7 1800 Therefore x=∑fiix==45N 40 i=1 17 1 xfxx and M.D. ()= ∑i =×400 =10 i− N 40 i=1 Shortcut method for calculating mean deviation about mean We can avoid the tedious calculations of computing xby following step-deviation method. Recall that in this method, we take an assumed mean which is in the middle or just close to it in the data. Then deviations of the observations (or mid-points of classes) are taken from the assumed mean. This is nothing but the shifting of origin from zero to the assumed mean on the number line, as shown in Fig 15.3 Fig 15.3 If there is a common factor of all the deviations, we divide them by this common factor to further simplify the deviations. These are known as step-deviations. The process of taking step-deviations is the change of scale on the number line as shown in Fig 15.4 Fig 15.4 The deviations and step-deviations reduce the size of the observations, so that the computations viz. multiplication, etc., become simpler. Let, the new variable be denoted xi−aby di = , where ‘a’ is the assumed mean and h is the common factor. Then, the h mean xby step-deviation method is given by n ∑fdiii=1x=+a ×h N Let us take the data of Example 6 and find the mean deviation by using step-deviation method. Take the assumed mean a= 45 and h = 10, and form the following Table 15.5. Table 15.5 Marks obtained Number of students Mid-points 45 10 i i xd − = iifd x xi− fi x xi− fi xi 10-20 2 15 – 3 – 6 30 60 20-30 3 25 – 2 – 6 20 60 30-40 8 35 – 1 – 8 10 80 40-50 14 45 0 0 0 0 50-60 8 55 1 8 10 80 60-70 3 65 2 6 20 60 70-80 2 75 3 6 30 60 40 0 400 7 ∑fdiiTherefore i=1xa ×h=+ N 0 = 45 +×10 =45 40 17 400 and .. x= fxx ==10 MD () ∑i i− N 40 i=1 $Note The step deviation method is applied to compute x. Rest of the procedure is same. (ii) Mean deviation about median The process of finding the mean deviation about median for a continuous frequency distribution is similar as we did for mean deviation about the mean. The only difference lies in the replacement of the mean by median while taking deviations. Let us recall the process of finding median for a continuous frequency distribution. The data is first arranged in ascending order. Then, the median of continuous frequency distribution is obtained by first identifying the class in which median lies (median class) and then applying the formula N −C2Median =+l ×hf where median class is the class interval whose cumulative frequency is just greater Nthan or equal to , N is the sum of frequencies, l, f, h and C are, respectively the lower 2 limit , the frequency, the width of the median class and C the cumulative frequency of the class just preceding the median class. After finding the median, the absolute values of the deviations of mid-point xi of each class from the median i.e., xi −M are obtained. 1 n Then M.D. (M) =∑fx−MiiN i=1 The process is illustrated in the following example: Example 7 Calculate the mean deviation about median for the following data : Class 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 6 7 15 16 4 2 Solution Form the following Table 15.6 from the given data : Table 15.6 N th The class interval containing or 25th item is 20-30. Therefore, 20–30 is the median2 class. We know that N − C2Median = l+× hf Here l= 20, C = 13, f= 15, h= 10 and N = 50 25 −13 Therefore, Median = 20 +×10 = 20 + 8 = 28 15 Thus, Mean deviation about median is given by fxi− MM.D. (M) = 1 ∑6 i 1 = ×508 = 10.1650N i=1 EXERCISE 15.1 Find the mean deviation about the mean for the data in Exercises 1 and 2. 1. 4, 7, 8, 9, 10, 12, 13, 17 2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44 Find the mean deviation about the median for the data in Exercises 3 and 4. 3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49 Find the mean deviation about the mean for the data in Exercises 5 and 6. 5. xi 5 10 15 20 25 fi 74635 6. xi 10 30 50 7090 fi 4 24 28 16 8 Find the mean deviation about the median for the data in Exercises 7 and 8. 7. xi 5 7 9 10 12 15 fi 8 6 2 2 2 6 8. xi 15 21 27 30 35 fi 3 5 6 7 8 Find the mean deviation about the mean for the data in Exercises 9 and 10. 9. Income 0-100 100-200 200-300 300-400 400-500 500-600 600-700 700-800 per day Number 4 8 9 107 5 4 3 of persons 10. Height 95-105 105-115 115-125 125-135 135-145 145-155 in cms Number of 9 13 26 30 12 10 boys 11. Find the mean deviation about median for the following data : Marks 0-10 10-20 20-30 30-40 40-50 50-60 Number of6 8 14 16 4 2 Girls 12. Calculate the mean deviation about median age for the age distribution of 100 persons given below: Age 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 Number5 6 12 14 26 12 16 9 [Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval] 15.4.3 Limitations of mean deviation In a series, where the degree of variability is very high, the median is not a representative central tendency. Thus, the mean deviation about median calculated for such series can not be fully relied. The sum of the deviations from the mean (minus signs ignored) is more than the sum of the deviations from median. Therefore, the mean deviation about the mean is not very scientific.Thus, in many cases, mean deviation may give unsatisfactory results. Also mean deviation is calculated on the basis of absolute values of the deviations and therefore, cannot be subjected to further algebraic treatment. This implies that we must have some other measure of dispersion. Standard deviation is such a measure of dispersion. 15.5 Variance and Standard Deviation Recall that while calculating mean deviation about mean or median, the absolute values of the deviations were taken. The absolute values were taken to give meaning to the mean deviation, otherwise the deviations may cancel among themselves. Another way to overcome this difficulty which arose due to the signs of deviations, is to take squares of all the deviations. Obviously all these squares of deviations are non-negative. Let x1, x2, x3, ..., xn be n observations and x be their mean. Then n 22 22(x1 x)(x2 x) ....... (xn x)(xi x) . i 1 If this sum is zero, then each (xi − x) has to be zero. This implies that there is no dispersion at all as all observations are equal to the mean x . n )2xIf ∑(xi − is small , this indicates that the observations x1, x2, x3,...,xn are i=1 x and therefore, there is a lower degree of dispersion. On the contrary, if this sum is large, there is a higher degree of dispersion of the observations close to the mean n from the mean x . Can we thus say that the sum ∑(xi − x)2 is a reasonable indicator i=1 of the degree of dispersion or scatter? Let us take the set A of six observations 5, 15, 25, 35, 45, 55. The mean of the observations is x = 30. The sum of squares of deviations from x for this set is 6 ∑(xi − x)2 = (5–30)2 + (15–30)2 + (25–30)2 + (35–30)2 + (45–30)2 +(55–30)2i=1 = 625 + 225 + 25 + 25 + 225 + 625 = 1750 Let us now take another set B of 31 observations 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45. The mean of these observations is y = 30 Note that both the sets A and B of observations have a mean of 30. Now, the sum of squares of deviations of observations for set B from the mean y is given by 31 )2∑( yi − y = (15–30)2 +(16–30)2 + (17–30)2 + ...+ (44–30)2 +(45–30)2 i=1 = (–15)2 +(–14)2 + ...+ (–1)2 + 02 + 12 + 22 + 32 + ...+ 142 + 152 = 2 [152 + 142 + ... + 12] 15 × (15 +1) (30 +1) = 2 × = 5 × 16 × 31 = 24806 ( +1) (2 nnn +1) (Because sum of squares of first n natural numbers = . Here n = 15)6 n )2If ∑(xi − x is simply our measure of dispersion or scatter about mean, we i=1 will tend to say that the set A of six observations has a lesser dispersion about the mean than the set B of 31 observations, even though the observations in set A are more scattered from the mean (the range of deviations being from –25 to 25) than in the set B (where the range of deviations is from –15 to 15). This is also clear from the following diagrams. For the set A, we have Fig 15.5 For the set B, we have Fig 15.6 Thus, we can say that the sum of squares of deviations from the mean is not a proper measure of dispersion. To overcome this difficulty we take the mean of the squares of the deviations, i.e., we take 1 ∑ n (xi − x)2 . In case of the set A, we have n i=1 11Mean = × 1750 = 291.67 and in case of the set B, it is × 2480 = 80.6 31 This indicates that the scatter or dispersion is more in set A than the scatter or dispersion in set B, which confirms with the geometrical representation of the two sets. 1 )2 Thus, we can take ∑(xi − x as a quantity which leads to a proper measuren of dispersion. This number, i.e., mean of the squares of the deviations from mean is called the variance and is denoted by σ 2 (read as sigma square). Therefore, the variance of n observations x1, x2,..., xn is given by n 21 2σ=∑(xi − x)n i=1 15.5.1 Standard Deviation In the calculation of variance, we find that the units of individual observations xi and the unit of their mean x are different from that of variance, since variance involves the sum of squares of (xi– x ). For this reason, the proper measure of dispersion about the mean of a set of observations is expressed as positive square-root of the variance and is called standard deviation. Therefore, the standard deviation, usually denoted by σ , is given by σ= 1 ∑ n (xi − x)2 ... (1)n i=1 Let us take the following example to illustrate the calculation of variance and hence, standard deviation of ungrouped data. Example 8 Find the Variance of the following data: 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 Solution From the given data we can form the following Table 15.7. The mean is calculated by step-deviation method taking 14 as assumed mean. The number of observations is n = 10 Table 15.7 xi 14 2 i i xd − = Deviations from mean (xi – x ) (xi – x ) 6 8 10 12 14 16 18 20 22 24 –4 –3 –2 –1 0 1 2 3 4 5 –9 –7 –5 –3 –1 1 3 5 7 9 81 49 25 9 1 1 9 25 49 81 5 330 ∑ n di i=15 ×= 15Therefore Mean x= assumed mean + ×h = 14 + 2 n 10 110 21and Variance ( σ2) = ∑(xi −x)= ×330 = 33ni=1 10 Thus Standard deviation (σ) = 33 574 =. 15.5.2 Standard deviation of a discrete frequency distribution Let the given discrete frequency distribution be x: x1, x2, x3 ,. . . , xn f: f1, f2, f3 ,. . . , fn 1 n 2In this case standard deviation ()σ= ∑fi(xi − x) ... (2)N i=1 n where N =∑fi . i=1 Let us take up following example. Example 9 Find the variance and standard deviation for the following data: xi 4 8 11 17 20 24 32 fi 3 5 9 5 4 3 1 Solution Presenting the data in tabular form (Table 15.8), we get Table 15.8 xi fi fi xi xi – x )2( xxi − fi )2( xxi − 4 8 11 17 20 24 32 3 5 9 5 4 3 1 12 40 99 85 80 72 32 –10 –6 –3 3 6 10 18 100 36 9 9 36 100 324 300 180 81 45 144 300 324 30 420 1374 77 N = 30, ∑ ii = 420, ∑ i( i−)2 =1374 fx fxx i=1 i=1 7 ∑ fxii 1Therefore i=1x= =× 420 =14 N 30 2 f(x− x)2Hence variance (σ ) = 1 ∑ 7 ii N i=1 1 = × 1374 = 45.830and Standard deviation (σ) = 45.8 = 6.77 15.5.3 Standard deviation of a continuous frequency distribution The given continuous frequency distribution can be represented as a discrete frequency distribution by replacing each class by its mid-point. Then, the standard deviation is calculated by the technique adopted in the case of a discrete frequency distribution. If there is a frequency distribution of nclasses each class defined by its mid-point xiwith frequency fi, the standard deviation will be obtained by the formula 2σ= 1 ∑ n fi(xi− x),N i=1 n where xis the mean of the distribution and N =∑ fi. i=1 Another formula for standard deviation We know that 21 n 21 n 22Variance = ∑ fi(xi− x)= ∑ fi(xi+ x− 2xxi)()σN i=1N i=1 n nn1 ⎡ 22 ⎤ = ⎢∑ fiix+∑xfi−∑2xfiix⎥N ⎣ i=1 i=1 i=1 ⎦ n nn1 ⎡ 22 ⎤ = ⎢∑ fiix+ x∑ fi− 2x∑xii f⎥N ⎣ i=1 i=1 i=1 ⎦ ⎡ 1 nn ⎤ = 1 ⎡n fx +x2N2− x.N x ⎤ Here ∑xf =xor ∑xf =Nx⎢∑ii ⎥⎢ ii ii ⎥N ⎣i=1 ⎦⎣ N i=1 i=1 ⎦ 1 22 21 22 = fx +x −2x = fx − x∑ n ii N ∑ n N i=1 i=1 ii n fx n ⎜⎛∑ii ⎟⎞2 ⎡ nn 2 ⎤11 ⎛⎞2 ⎜i=1 ⎟ 2 or σ2 = fx − =⎢N fx −⎜ fx ⎥∑ii ∑ii ∑ii ⎟⎜⎟ 2N i−1 NN ⎢ i=1 ⎝i=1 ⎠⎥⎣⎦⎜⎟⎝⎠ 12 ⎛⎞σ= fxThus, standard deviation () N∑ n −∑n fx 2 ... (3)ii ⎜ ii ⎟N i=1 ⎝i=1 ⎠ Example 10 Calculate the mean, variance and standard deviation for the following distribution : Class 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 3 7 12 15 8 3 2 Solution From the given data, we construct the following Table 15.9. Table 15.9 Class Frequency (fi) Mid-point (xi) fixi (xi – x)2 fi(xi – x)2 30-40 40-50 50-60 60-70 70-80 80-90 90-100 3 7 12 15 8 3 2 35 45 55 65 75 85 95 105 315 660 975 600 255 190 729 289 49 9 169 529 1089 2187 2023 588 135 1352 1587 2178 50 3100 10050 1 7 3100 Thus Mean x=∑ fiix== 62 N i=1 50 17 2σ 2f(xxVariance () = ∑ ii − )N i=1 1 = ×10050 = 20150 and Standard deviation (σ) = 201 =14 18. Example 11 Find the standard deviation for the following data : xi 3 8 13 18 23 fi 7 10 15 10 6 Solution Let us form the following Table 15.10: Table 15.10 2xxi fi fixi i fixi 2 3 7 21 9 63 8 10 80 64 640 13 15 195 169 2535 18 10 180 324 3240 23 6 138 529 3174 48 614 9652 Now, by formula (3), we have 1 N fx− fxσ = ∑ ii 2 (∑ ii)2 N =1 48 × 9652 − (614) 2 48 1 = 463296 − 376996 48 1 .= ×293 77 = 6.1248 Therefore, Standard deviation (σ ) = 6.12 15.5.4. Shortcut method to find variance and standard deviation Sometimes the values of xi in a discrete distribution or the mid points xi of different classes in a continuous distribution are large and so the calculation of mean and variance becomes tedious and time consuming. By using step-deviation method, it is possible to simplify the procedure. 1Let the assumed mean be ‘A’ and the scale be reduced to times (hbeing theh width of class-intervals). Let the step-deviations or the new values be yi. xi − A ih i.e. y = or xi = A + hyi ... (1) n fx∑ ii i=1We know that x= ... (2)N Replacing xi from (1) in (2), we get n ∑ fi(A + hyi) x = i=1 N n1 n 1 ⎛ nn ⎞ fA + hfy A f+h fy = ⎛⎜∑ i ∑ ii ⎞⎟ = ⎜∑ i ∑ ii⎟N ⎝ i=1 i=1 ⎠ N ⎝ i=1 i=1 ⎠ n nfy ⎛∑ ii ⎞N ⎜ because ∑ fi = N ⎟= i=1A . + h ⎝ i=1 ⎠NN Thus x=A + hy ... (3) n 21 2Now Variance of the variable x, σ x = N ∑ fi(xi− x) i=1 1 n 2 = ∑ fi(A + hyi− A − hy) (Using (1) and (3))N i=1 n12 2 = ∑ i ( i− )fhy yN i=1 h2 n 2fy y=N ∑ i( i− )= h2 × variance of the variable yii=1 i.e. σ x 2= h2σ 2 y or = hσ ... (4)σ xy From (3) and (4), we have nnh 2 ⎛⎞2 σ x =N N∑ fiiy−⎜∑ fiiy⎟ ... (5) i=1 ⎝ i=1 ⎠ Let us solve Example 11 by the short-cut method and using formula (5) Examples 12 Calculate mean, Variance and Standard Deviation for the following distribution. Classes 30-40 40-50 50-60 60-70 70-80 80-90 90-100 Frequency 3 7 12 15 8 3 2 Solution Let the assumed mean A = 65. Here h= 10 We obtain the following Table 15.11 from the given data : Table 15.11 Class Frequency Mid-point yi = 65 10 ix− yi 2 fiyi fiyi 2 fi xi 30-40 3 35 – 3 9 – 9 27 40-50 7 45 – 2 4 – 14 28 50-60 12 55 – 1 1 – 12 12 60-70 15 65 0 0 0 0 70-80 8 75 1 1 8 8 80-90 3 85 2 4 6 12 9 0-100 2 95 3 9 6 18 N=50 – 15 105 fy∑ii15×= 65 −=62 Therefore x= A + h ×10 50 50 2h ⎡ 2 ⎤Variance σ2 = ⎢N fy2 −(∑ii ⎥∑ii fy)N2 ⎣⎦ 10()22 ⎤⎡50 ×105 −(–15) = ⎣⎦(50)2 ⎢⎥ 1 = [5250 −225] =201 25 and standard deviation () 201 = 14.18σ=EXERCISE 15.2 Find the mean and variance for each of the data in Exercies 1 to 5.1. 6, 7, 10, 12, 13, 4, 8, 122. First nnatural numbers3. First 10 multiples of 34. 5. 6. Find the mean and standard deviation using short-cut method. Find the mean and variance for the following frequency distributions in Exercises 7 and 8.7. Classes 0-30 30-60 60-90 90-120 120-150 150-180 180-210 Frequencies 2 3 5 10 3 5 2 8. Classes 0-10 10-20 20-30 30-40 40-50 Frequencies 5 8 15 16 6 9. Find the mean, variance and standard deviation using short-cut method 10. The diameters of circles (in mm) drawn in a design are given below: Height in cms 70-75 75-80 80-85 85-90 90-95 95-100 100-105 105-110 110-115 No. of children 3 4 7 7 15 9 6 6 3 Diameters 33-36 37-40 41-44 45-48 49-52 No. of circles 15 17 21 22 25 Calculate the standard deviation and mean diameter of the circles. [ Hint First make the data continuous by making the classes as 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 - 48.5, 48.5 - 52.5 and then proceed.] 15.6 Analysis of Frequency Distributions In earlier sections, we have studied about some types of measures of dispersion. The mean deviation and the standard deviation have the same units in which the data are given. Whenever we want to compare the variability of two series with same mean, which are measured in different units, we do not merely calculate the measures of dispersion but we require such measures which are independent of the units. The measure of variability which is independent of units is called coefficient of variation (denoted as C.V.) The coefficient of variation is defined as σC.V. =×100, x ≠0, x where σ and x are the standard deviation and mean of the data. For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other. 15.6.1 Comparison of two frequency distributions with same mean Let x1 and σ1 be the mean and standard deviation of the first distribution, and x2 and σ2 be the mean and standard deviation of the second distribution. σ1 ×100Then C.V. (1st distribution) = x1 σ2 ×100 and C.V. (2nd distribution) = x2 Given x1= x2 = x (say) σ1Therefore C.V. (1st distribution) = ×100 ... (1)x σand C.V. (2nd distribution) = 2 ×100 ... (2)x It is clear from (1) and (2) that the two C.Vs. can be compared on the basis of values of and only. σ1 σ2 Thus, we say that for two series with equal means, the series with greater standard deviation (or variance) is called more variable or dispersed than the other. Also, the series with lesser value of standard deviation (or variance) is said to be more consistent than the other. Let us now take following examples: Example 13 Two plants A and B of a factory show following results about the number of workers and the wages paid to them. AB No. of workers 5000 6000 Average monthly wages Rs 2500 Rs 2500 Variance of distribution 81 100 of wages In which plant, A or B is there greater variability in individual wages? Solution The variance of the distribution of wages in plant A ( σ12 ) = 81 Therefore, standard deviation of the distribution of wages in plant A ( σ1) = 9 Also, the variance of the distribution of wages in plant B ( 2 ) = 100σ2 Therefore, standard deviation of the distribution of wages in plant B (σ2) = 10 Since the average monthly wages in both the plants is same, i.e., Rs.2500, therefore, the plant with greater standard deviation will have more variability. Thus, the plant B has greater variability in the individual wages. Example 14 Coefficient of variation of two distributions are 60 and 70, and their standard deviations are 21 and 16, respectively. What are their arithmetic means. Solution Given C.V. (1st distribution) = 60, σ1= 21 C.V. (2nd distribution) = 70, σ2 = 16 Let x1 and x2 be the means of 1st and 2nd distribution, respectively. Then σ1C.V. (1st distribution) = x × 100 1 21 21 Therefore 60 = ×100 or x =×100 = 35 x1 160 σ 2and C.V. (2nd distribution) = x ×100 2 16 16i.e. 70 = ×100 or x2 =×100 = 22 85 . x2 70 Example 15 The following values are calculated in respect of heights and weights of the students of a section of Class XI : Height Weight Mean 162.6 cm 52.36 kg Variance 127.69 cm2 23.1361 kg2 Can we say that the weights show greater variation than the heights? Solution To compare the variability, we have to calculate their coefficients of variation. Given Variance of height = 127.69cm2 Therefore Standard deviation of height = 127.69cm = 11.3 cm Also Variance of weight = 23.1361 kg2 Therefore Standard deviation of weight = .23 1361 kg= 4.81 kg Now, the coefficient of variations (C.V.) are given by (C.V.) in heights = Standard Deviation 100 × Mean = 11 3 100 162 6 . . × = 6.95 and (C.V.) in weights = 481 52 36 . . 100 × = 9.18 Clearly C.V. in weights is greater than the C.V. in heights Therefore, we can say that weights show more variability than heights. EXERCISE 15.3 1. From the data given below state which group is more variable, A or B? Marks 10-20 20-30 30-40 40-50 50-60 60-70 70-80 Group A 9 17 32 33 40 10 9 Group B 10 20 30 25 43 15 7 2. From the prices of shares X and Y below, find out which is more stable in value: 3. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results: Firm A Firm B No. of wage earners 586 648 Mean of monthly wages Rs 5253 Rs 5253 Variance of the distribution 100 121 of wages (i) Which firm A or B pays larger amount as monthly wages? (ii) Which firm, A or B, shows greater variability in individual wages? 4. The following is the record of goals scored by team A in a football session: For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent? 5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below: 50 50 5050 ∑ i , ∑ i 2 902 8 , ∑ i , ∑ i 2 1457 6 x = 212 x = .y = 261 y = . i=1 i=1 i=1 i=1 Which is more varying, the length or weight? Miscellaneous Examples Example 16 The variance of 20 observations is 5. If each observation is multiplied by 2, find the new variance of the resulting observations. Solution Let the observations be x1, x2, ..., x20 and x be their mean. Given that variance = 5 and n = 20. We know that 20 12012 2Variance (σ 2 )= ∑(xi − x) , i.e., 5 =∑(xi − x)20 =n i 1i=1 20 or ∑(xi − x )2 = 100 ... (1) i=1 If each observation is multiplied by 2, and the new resulting observations are yi , then 1 = 2x i.e., x = yiyiii2 2020 2011 1 Therefore y =∑ yi =∑2xi = 2 . ∑xi n 20 = 20 =i=1 i 1 i 1 1 i.e. y = 2 x or x = y2 Substituting the values of xi and x in (1), we get STATISTICS 377 20 ⎛11 ⎞2 202∑⎜yi − y ⎟=100 , i.e., ∑( yi −y) =400 i=1 ⎝22 ⎠ i=1 12Thus the variance of new observations = ×400 =20 =2 ×5 20 $Note The reader may note that if each observation is multiplied by a constant k, the variance of the resulting observations becomes k2 times the original variance. Example17 The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations. Solution Let the other two observations be x and y. Therefore, the series is 1, 2, 6, x, y. 126 xy++++ Now Mean x = 4.4 = 5 or 22 =9 + x + y Therefore x + y = 13 ... (1) 15 2Also variance = 8.24 = ∑(xi −x)ni=1 12 2222 2 ⎤i.e. 8.24 = ⎡()() () 34 +24 +16 + + −× xy 2 44( .xy)2 () .. .. + +× 44⎣⎦5 or 41.20 = 11.56 + 5.76 + 2.56 + x2 + y2 –8.8 × 13 + 38.72 Therefore x2 + y2 = 97 ... (2) But from (1), we have x2 + y2 + 2xy = 169 ... (3) From (2) and (3), we have 2xy = 72 ... (4) Subtracting (4) from (2), we get x2 + y2 – 2xy = 97 – 72 i.e. (x – y)2 = 25 or x – y = ±5 ... (5) So, from (1) and (5), we get x = 9, y = 4 when x – y = 5 or x = 4, y = 9 when x – y = – 5 Thus, the remaining observations are 4 and 9. Example 18 If each of the observation x1, x2, ...,xn is increased by ‘a’, where a is a negative or positive number, show that the variance remains unchanged. Solution Let x be the mean of x1, x2, ...,xn . Then the variance is given by σ12 = 1 ∑ n (xi − x)2 n i=1 If ‘a is added to each observation, the new observations will be yi = xi + a ... (1) Let the mean of the new observations be y . Then 1 n 1 n y = n ∑ yi = n ∑(xi + a) i=1 i=1 nn n1 ⎡⎤ 1 na = ⎢∑xi +∑a⎥ = ∑xi += x + a n ⎣ i=1 i=1 ⎦ ni=1 n i.e. y = x + a ... (2) Thus, the variance of the new observations 1 n 21 n σ22 = ∑(yi − y)= ∑(xi + a − x − a)2 [Using (1) and (2)]nni=1 i=1 1 n 2∑ (x − x)2= i = n σ1i=1 Thus, the variance of the new observations is same as that of the original observations. $Note We may note that adding (or subtracting) a positive number to (or from) each observation of a group does not affect the variance. Example 19 The mean and standard deviation of 100 observations were calculated as 40 and 5.1, respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation? Solution Given that number of observations (n) = 100 Incorrect mean ( x ) = 40, Incorrect standard deviation (σ) = 5.1 n We know that x = 1 ∑ xin i=1 100 100 1 i.e. 40 = 100 ∑xi or ∑xi = 4000 i=1 i=1 i.e. Incorrect sum of observations = 4000 Thus the correct sum of observations = Incorrect sum – 50 + 40 = 4000 – 50 + 40 = 3990 correct sum 3990 Hence Correct mean = = = 39.9100 100 1 n 21 ⎛n ⎞2 Also Standard deviation σ= ∑xi − 2 ⎜∑xi ⎟ nni=1 ⎝i=1 ⎠ 1 22x= ∑ n xi −()n i=1 n1 22i.e. 5.1 = ×Incorrect ∑xi −(40) 100 i =1 1 n 2 or 26.01 = ×Incorrect ∑x – 1600100 i =1 i Therefore Incorrect ∑ n xi 2 = 100 (26.01 + 1600) = 162601 i=1 n n xNow Correct ∑i 2 = Incorrect ∑xi 2 – (50)2 + (40)2 i=1 i=1 = 162601 – 2500 + 1600 = 161701 Therefore Correct standard deviation Miscellaneous Exercise On Chapter 15 1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations. 2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations. 3. The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations. 4. Given that x is the mean and σ2 is the variance of n observations x1, x2, ...,xn . Prove that the mean and variance of the observations ax1, ax2, ax3, ...., axn are ax and a2 σ2, respectively, (a ≠ 0). 5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases: (i) If wrong item is omitted. (ii) If it is replaced by 12. 6. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below: Subject Mathematics Physics Chemistry Mean 42 32 40.9 Standard 12 15 20 deviation which of the three subjects shows the highest variability in marks and which shows the lowest? 7. The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted. Summary Measures of dispersion Range, Quartile deviation, mean deviation, variance, standard deviation are measures of dispersion. Range = Maximum Value – Minimum Value Mean deviation for ungrouped data x–x xi – M∑M.D.( ) = i, M.D. (M) =x ∑ n n Mean deviation for grouped data fxi – x fxi –M ∑ i ∑ iM.D.( )x = , M.D. (M) = , where N =∑ f NN i Variance and standard deviation for ungrouped data 2121 2σ= (xi –x ) σ= (– x)∑ , ∑ xi nn Variance and standard deviation of a discrete frequency distribution 21 212σ= ∑ fi ( xi − x ) , σ= ∑ fi ( xi − x )NN Variance and standard deviation of a continuous frequency distribution σ 2 = 1 f (x − x )2, σ= 1N fx2 − fx )2 N ∑ ii N ∑ ii (∑ ii Shortcut method to find variance and standard deviation. 2 2 h ⎡ 22 ⎤ h 2σ= N fy −( fy ) , σ= N fy −( fy )2,N2 ⎢⎣ ∑ ii ∑ ii ⎥⎦ N ∑ ii ∑ ii xi − Awhere yi = h σCoefficient of variation (C.V.) =×100, x ≠ 0. x For series with equal means, the series with lesser standard deviation is more consistent or less scattered. Historical Note ‘Statistics’ is derived from the Latin word ‘status’ which means a political state. This suggests that statistics is as old as human civilisation. In the year 3050 B.C., perhaps the first census was held in Egypt. In India also, about 2000 years ago, we had an efficient system of collecting administrative statistics, particularly, during the regime of Chandra Gupta Maurya (324-300 B.C.). The system of collecting data related to births and deaths is mentioned in Kautilya’s Arthshastra (around 300 B.C.) A detailed account of administrative surveys conducted during Akbar’s regime is given in Ain-I-Akbari written by Abul Fazl. Captain John Graunt of London (1620-1674) is known as father of vital statistics due to his studies on statistics of births and deaths. Jacob Bernoulli (1654-1705) stated the Law of Large numbers in his book “Ars Conjectandi’, published in 1713. The theoretical development of statistics came during the mid seventeenth century and continued after that with the introduction of theory of games and chance (i.e., probability). Francis Galton (1822-1921), an Englishman, pioneered the use of statistical methods, in the field of Biometry. Karl Pearson (1857-1936) contributed a lot to the development of statistical studies with his discovery of Chi square test and foundation of statistical laboratory in England (1911). Sir Ronald A. Fisher (1890-1962), known as the Father of modern statistics, applied it to various diversified fields such as Genetics, Biometry, Education, Agriculture, etc. — • —

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