Edusat  Session 4  Nagesh  S J C E
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Edusat Session 4 Nagesh S J C E

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    Edusat  Session 4  Nagesh  S J C E Edusat Session 4 Nagesh S J C E Document Transcript

    • Session – 4 Measures of Central Tendency A classified statistical data may sometimes be described as distributed around some value called the central value or average is some sense. It gives the most representative value of the entire data. Different methods give different central values and are referred to as the measures of central tendency. Thus, the most important objective of statistical analysis is to determine a single value that represents the characteristics of the entire raw data. This single value representing the entire data is called ‘Central value’ or an ‘average’. This value is the point around which all other values of data cluster. Therefore, it is known as the measure of location and since this value is located at central point nearest to other values of the data it is also called as measures of central tendency. Different methods give different central values and are referred as measures of central tendency. The common measures of central tendency are a) Mean b) Median c) Mode. These values are very useful not only in presenting overall picture of entire data, but also for the purpose of making comparison among two or more sets of data. Average Definition Average is a value which is typical or representative of a set of data. - Murry R. Speigal Average is an attempt to find one single figure to describe whole of figures. - Clark & Sekkade From above definitions it is clear that average is a typical value of the entire data and is a measure of central tendency. Functions of an average • To represents complex or large data. • It facilitates comparative study of two variables. • Helps to study population from sample data. • Helps in decision making. • Represents single value for a series of data. • To establish mathematical relationship. 1
    • Characteristics of a typical average • It should be rigidly defined and easily understandable. • It should be simple to compute and in the form of mathematical formula. • It should be based on all the items in the data. • It should not be unduly influenced by any single item. • It should be capable of further mathematical treatment. • It should have sampling stability. Types of average Average or measures of central tendency are of following types. 1. Mathematical average a. Arithmetical mean i. Simple mean ii. Weighted mean b. Geometric mean c. Harmonic mean 2. Positional Averages a. Median b. Mode Arithmetic mean Arithmetic mean is also called arithmetic average. It is most commonly used measures of central tendency. Arithmetic average of a series is the value obtained by dividing the total value of various item by its number. Arithmetic average are of two types a. Simple arithmetic average b. Weighted arithmetic average Simple arithmetic average (Mean) Arithmetic mean is simply sometimes referred as ‘Mean’. Ex: Mean income, Mean expenses, Mean marks etc. Unlike other averages, mean has to be computed by considering each and every observations in the series. Hence, the mean cannot be found by either by inspection or observation of items. Simple arithmetic mean is equal to sum of the variable divided by their number of observations in the sample. 2
    • Let xi is the variable which takes values x1, x2, x3,……… xn over ‘n’ items, then arithmetic mean, simply the mean of x, denoted by bar over the variable x is given by. x= x 1 + x 2 + x 3 + ............... + x n = ∑x n n Where, Σ is the Greek symbol sigma denotes the summation of all xi values. Arithmetic mean can be computed by following two methods for direct observation of individual items. a. Direct method b. Short cut method. Direct method uses above equation and steps for short cut method is illustrated in the subsequent topic. Ex: (For Direct Method) 1. Calculate the mean for following data. Marks obtained by 65 students are given below: 20, 15, 23, 22, 25, 20. x 1 + x 2 + ......... + x n Mean marks x= n 20 + 15 + 23 + 22 + 25 + 20 = 6 125 = 6 = 20.83 2. Six month income of departmental store are given below. Find mean income of stores. Month Jan Feb Mar Apr May June Income (Rs.) 25000 30000 45000 20000 25000 20000 n = Total No. of items (observations) = 6 Total income = Σxi = (25000 + 30000 + 45000 + 20000 + 20000) = 140000 Mean income = ∑x i = 140000 = Rs. 23333.33 n 6 The above example shows that if there are large data or large figures are there in data, computations required to get mean in high. In order to reduce computations one can go for short-cut method. The method is illustrated below. 3
    • Shortcut method Steps of this method is given below. Step 1: Assume any one value as a mean which is called arbitrary average (A). Step 2: Find the difference (deviations) of each value from arbitrary average. D = xi – A Step 3: Add all deviations (differences) to get Σd. Step 4: Use following equation and compute the mean value. x=A+ ∑d n n = Total No. of observations Σd = Total deviation value A = Arbitrary mean Example: Find the mean marks obtained by the students for the joining data given. 20 25 20 22 20 21 23 25 22 18 Let A = 20 and n = 10 Marks D = (xi – 20) 20 0 25 5 20 0 22 2 20 0 21 1 23 3 25 5 22 2 18 -2 Σd = 16 x=A+ ∑d n 16 x = 20 + 10 = 20 + 1.6 Mean Marks x = 21.6 4
    • 1. Mathematical characteristics of mean a. Algebraic sum of deviations of all observations from their arithmetic mean is zero i.e. Σ(xi - x ) = 0. b. The sum of squared deviations of the items from the mean is a minimum, that is less than the sum of squared deviations of items from any other value. Σd2 = minimum c. Since x = ∑ . If any two values are given, third value can be computed. x n d. If all the items of a sets are increased / decreased by any constant value, the arithmetic mean will also increases / decreases by the same constant. 2. Weighted arithmetic mean The weighted mean is computed by considering the relative importance of each of values to the total value. The arithmetic mean gives equal importance to all the items of distribution. In certain cases, relative importance of items is not the same. To give relative importance, weightage may be given to variables depending on cases. Thus, weightage represents the relative importance of the items. The weighted arithmetic mean in computed by following equation. Let x1, x2, x3, ………… xn are the variables and w1, w2, w3, ………… wn are the respective weights assigned. Then weighted mean x w is given by below equation. xw = x 1 w 1 + x 2 w 2 + x 3 w 3 + ...... + x n w n = ∑ xw w 1 + w 2 + w 3 + ............ + w n ∑w i.e., weighted average is the ratio of product of all values and respective weights to sum of weights. Ex: Compute simple weighted arithmetic mean and comment on them. Monthly salary Strength of Designation xw (Rs) (x) cadre (w) General Manager 25000 10 250000 Mangers 19000 20 380000 Supervisors 14000 10 140000 Office Assistant 10000 50 500000 Helpers 8000 25 200000 (N = 5) Total Σx = 76000 Σw = 115 Σxw = 1470000 5
    • ∑ x 76000 a. Simple arithmetic mean = = = Rs. 15200 N 5 ∑ xw 1470000 b. Weighted arithmetic mean = = = Rs. 12782.6 ∑w 115 In this example, simple arithmetic mean does not accounts the difference in salary range for various staff. It is given equal importance. The salary of General Manager and Manager has inflated the value of simple mean. The weighted mean gives importance to the number of persons in various salary range. Ex: Comment on performance of students of two universities given below. Universit Bombay Madras y No. of (w) No. of % of % of Course students wx (w) wx pas (x) pas (x) (000) students MBA 71 3 213 81 5 405 MCA 83 2 166 76 3 228 MA 73 5 365 58 3 174 M.Sc. 75 2 150 76 1 76 M.Com. 70 2 140 81 2 162 Total (Σ) Σx = Σw =14 Σwx =1034 Σx =372 Σw =14 Σwx =1045 372 a. Since Σx is same, simple arithmetic average for both universities. ∑ x 372 = = = 74.4 N 5 ∑ wx 1034 b. Weighted mean for Bombay University = = = 73.86 ∑w 14 ∑ wx 1045 c. Weighted mean for Madras University = = = 74.64 ∑w 14 Comment: Madras University student’s performance is better than Bombay University students. Discrete Series Frequencies of each value is multiplied with respective size to get total number of items is discrete series and their total number of item is divided by total number of frequencies to obtain arithmetic mean. This can be done in two methods one by direct or by short cut method. 6
    • Ex: Calculate the mean for following data. Value (x) 1 2 3 4 5 Frequency (f) 10 15 10 9 5 Steps: 1. Multiply each size of item by frequency to get Σfx 2. Add all frequencies (Σf = N) ∑ fx ∑ fx 3. Use formula x = = to get mean value. ∑f N Solution: By direct method Value (x) Frequency (f) fx 1 10 10 2 15 30 3 10 30 4 9 36 5 5 25 Σf = 49 Σfx = 131 ∑ fx 131 x= = = 2.67 N 49 By short-cut method Let A = 3, (Assumed mean = 3) Value (x) Frequency (f) d = (x –A) fd 1 10 -2 -20 2 15 -1 -15 3 10 0 0 4 9 1 9 5 5 2 10 Σf = 49 Σfd = - 16 ∑ fx  − 16  x=A+ =3 +   = 2.67 N  49  7
    • Continuous series In continuous frequency distribution, the individual value of each item in the frequency distribution is not known. In a continuous series the mid points of various class intervals are written down to replace the class interval. In continuous series the mean can be calculated by any of the following methods. a. Direct method b. Short cut method c. Step deviation method a. Direct method Steps of their method are as follows 1. Find out the mid value of class group or class. 23 + 30 50 Ex: For a class interval 20-30, the mid value is = = 25 mid value 2 2 is denoted by ‘m’. 2. Multiply the mid value ‘m’ by frequency ‘f’ of each class and sum up to get Σfm. ∑ fm 3. Use x = where N = Σf formula to get mean value. N Ex: Compute the mean for following data. Age group No. of persons Mid point fm (CI) (f) ‘m’ 0 – 10 5 5 25 10 – 20 15 15 225 20 – 30 25 25 625 30 – 40 8 35 280 40 – 50 7 45 315 Total Σf = 60 = N Σfm = 1470 ∑ fm ∑ fm 1470 Mean age = = = = 245 ∑f N 60 x = 24.5 b. Short cut method Steps of above methods are described below. 1. Find the mid value of each class 2. Assume any of the mid value as arbitrary average (A). 3. Multiply the deviation (differences) ‘d’ by frequency ‘f’. 8
    • ∑ fd Using the formula x = A + find the mean value. N Ex: Find the mean age of patient visiting to hospital in a particular day using following data. Age group No. of patients Mid value d = (m – 25) fd CI (f) M 0 – 10 5 5 -20 -100 10 – 20 15 15 -10 -150 20 – 30 25 25 0 0 30 – 40 8 35 10 80 40 – 50 7 45 20 140 Total Σf = 60 = N Σfd = –30 Let Arbitrary average = A = 25 ∑ fd Mean age x = A+ N  − 30  1 x = 25 +   = 25 − = 24.5  60  2 x = 24.5 c. Step deviation method In this method, after finding deviation from arbitrary mean, it is divided by a common factor. Scaling down the deviation by a ‘step’ will reduce the calculation to minimum. The procedure of this method is described below. Steps of step deviation method 1. Find out the mid value ‘m’. 2. Select the arbitrary men ‘A’. 3. Find the deviation (d) of mid value of each from ‘A’. 4. Deviations ‘d’ are divided by a common factor –d'. 5. multiply d' of each class by frequency ‘f’ to get fd' and sum up for all classes to get Σfd'. ∑ fd' 6. Using the formula x = A + x C (where, C is a common factor) N calculate mean value. 9
    • Ex: Find the mean age of following data. No. of persons Mid value (d=m–A) d Age (CI) d'= fd' ‘f’ ‘m’ (d=m–25) 10 0 – 10 5 5 -20 -2 -10 10 – 20 15 15 -10 -1 -15 20 – 30 25 25 0 0 0 30 – 40 8 35 10 1 8 40 – 50 7 45 20 2 14 Total Σf=60=N Σfd'= -3 Let A = 25 and C = 10 ∑ fd' x=A+ xC N (−3) x = 25 + x 10 60 1 x = 25 − 2 x = 24.5 10