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# Edusat Session 5 Nagesh S J C E

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### Edusat Session 5 Nagesh S J C E

1. 1. Session – 5 Measures of Central Tendency Combined Mean Combined arithmetic mean can be computed if we know the mean and number of items in each groups of the data. The following equation is used to compute combined mean. Let x 1 & x 2 are the mean of first and second group of data containing N1 & N2 items respectively. N1 x 1 + N 2 x 2 Then, combined mean = x 12 = N1 + N 2 N1 x 1 + N 2 x 2 + N 3 x 3 If there are 3 groups then x 123 = N1 + N 2 + N 3 Ex - 1: a) Find the means for the entire group of workers for the following data. Group – 1 Group – 2 Mean wages 75 60 No. of workers 1000 1500 Given data: N1 = 1000 N2 = 1500 x 1 = 75 & x 2 = 60 N1 x 1 + N 2 x 2 Group Mean = x 12 = N1 + N 2 1000 x 75 + 1500 x 60 = 1000 + 1500 = x 12 = Rs. 66 Ex - 2: Compute mean for entire group. Medical examination No. examined Mean weight (pounds) A 50 113 B 60 120 C 90 115 1
2. 2. N1 x 1 + N 2 x 2 + N 3 x 3 Combined mean (grouped mean weight) = N1 + N 2 + N 3 (50 x 113 + 60 x 120 + 90 x 115) x 123 = (50 + 60 + 90) x 123 = Mean weight = 116 pounds Merits of Arithmetic Mean 1. It is simple and easy to compute. 2. It is rigidly defined. 3. It can be used for further calculation. 4. It is based on all observations in the series. 5. It helps for direct comparison. 6. It is more stable measure of central tendency (ideal average). Limitations / Demerits of Mean 1. It is unduly affected by extreme items. 2. It is sometimes un-realistic. 3. It may leads to confusion. 4. Suitable only for quantitative data (for variables). 5. It can not be located by graphical method or by observations. Geometric Mean (GM) The GM is nth root of product of quantities of the series. It is observed by multiplying the values of items together and extracting the root of the product corresponding to the number of items. Thus, square root of the products of two items and cube root of the products of the three items are the Geometric Mean. Usually, geometric mean is never larger than arithmetic mean. If there are zero and negative number in the series. If there are zeros and negative numbers in the series, the geometric means cannot be used logarithms can be used to find geometric mean to reduce large number and to save time. In the field of business management various problems often arise relating to average percentage rate of change over a period of time. In such cases, the arithmetic mean is not an appropriate average to employ, so, that we can use geometric mean in such case. GM are highly useful in the construction of index numbers. Geometric Mean (GM) = n x 1 x x 2 x ...........x x n When the number of items in the series is larger than 3, the process of computing GM is difficult. To over come this, a logarithm of each size is obtained. 2
3. 3. The log of all the value added up and divided by number of items. The antilog of quotient obtained is the required GM.  log1 + log 2 + ................ + log n   ∩ log x i  (GM) = Antilog  n  Anti log i ∑1 N    =  Merits of GM a. It is based on all the observations in the series. b. It is rigidly defined. c. It is best suited for averages and ratios. d. It is less affected by extreme values. e. It is useful for studying social and economics data. Demerits of GM a. It is not simple to understand. b. It requires computational skill. c. GM cannot be computed if any of item is zero or negative. d. It has restricted application. Ex - 1: a. Find the GM of data 2, 4, 8 x1 = 2, x2 = 4, x3 = 8 n=3 GM = n x 1 x x 2 x x 3 GM = 3 2 x 4 x 8 GM = 3 64 = 4 GM = 4 b. Find GM of data 2, 4, 8 using logarithms. Data: x1 = 2 x2 = 4 x3 = 8 N=3 3
4. 4. x log x 2 0.301 4 0.602 8 0.903 Σlogx = 1.806  ∑ log x  GM = Antilog    N  1.806  GM = Antilog   3   GM = Antilog (0.6020) = 3.9997 GM ≅ 4 Ex - 2: Compare the previous year the Over Head (OH) expenses which went up to 32% in year 2003, then increased by 40% in next year and 50% increase in the following year. Calculate average increase in over head expenses. Let 100% OH Expenses at base year Year OH Expenses (x) log x 2002 Base year – 2003 132 2.126 2004 140 2.146 2005 150 2.176 Σ log x = 6.448  ∑ log x  GM = Antilog    N   6.448  GM = Antilog   3   GM = 141.03 GM for discrete series GM for discrete series is given with usual notations as month: 4
5. 5.  ∩ log x i  GM = Antilog i ∑1 N  =  Ex - 3: Consider following time series for monthly sales of ABC company for 4 months. Find average rate of change per monthly sales. Month Sales I 10000 II 8000 III 12000 IV 15000 Let Base year = 100% sales. Solution: Sales Increase / Conversion Month Base year decrease log (x) (Rs) (x) %ge I 100% 10000 – – – II – 20% 8000 80 80 1.903 III + 50% 12000 130 130 2.113 IV + 25% 15000 155 155 2.190 Σlogx = 6.206  6.206  GM = Antilog  = 117.13  3   Average sales = 117.13 – 100 = 14.46% Ex - 4: Find GM for following data. Marks No. of students log x f log x (x) (f) 130 3 2.113 6.339 135 4 2.130 8.52 140 6 2.146 12.876 145 6 2.161 12.996 150 3 2.176 6.528 Σf = N = 22 Σ f log x =47.23 5
6. 6.  ∑ f log x  GM = Antilog    N   47.23  GM = Antilog   22   GM = 140.212 Geometric Mean for continuous series Steps: 1. Find mid value m and take log of m for each mid value. 2. Multiply log m with frequency ‘f’ of each class to get f log m and sum up to obtain Σ f log m. 3. Divide Σ f log m by N and take antilog to get GM. Ex: Find out GM for given data below Yield of wheat No. of farms Mid value in frequency log m f log m ‘m’ MT (f) 1 – 10 3 5.5 0.740 2.220 11 – 20 16 15.5 1.190 19.040 21 – 30 26 25.5 1.406 36.556 31 – 40 31 35.5 1.550 48.050 41 – 50 16 45.5 1.658 26.528 51 – 60 8 55.5 1.744 13.954 Σf = N = 100 Σ f log m = 146.348  ∑ f log m  GM = Antilog    N  146.348  GM = Antilog   100   GM = 29.07 Harmonic Mean It is the total number of items of a value divided by the sum of reciprocal of values of variable. It is a specified average which solves problems involving variables expressed in within ‘Time rates’ that vary according to time. 6
7. 7. Ex: Speed in km/hr, min/day, price/unit. Harmonic Mean (HM) is suitable only when time factor is variable and the act being performed remains constant. N HM = 1 ∑ x Merits of Harmonic Mean 1. It is based on all observations. 2. It is rigidly defined. 3. It is suitable in case of series having wide dispersion. 4. It is suitable for further mathematical treatment. Demerits of Harmonic Mean 1. It is not easy to compute. 2. Cannot used when one of the item is zero. 3. It cannot represent distribution. Ex: 1. The daily income of 05 families in a very rural village are given below. Compute HM. Family Income (x) Reciprocal (1/x) 1 85 0.0117 2 90 0.01111 3 70 0.0142 4 50 0.02 5 60 0.016 ∑1 x = 0.0738 N HM = ∑1 x 5 = = 67.72 0.0738 HM = 67.72 7
8. 8. 2. A man travel by a car for 3 days he covered 480 km each day. On the first day he drives for 10 hrs at the rate of 48 KMPH, on the second day for 12 hrs at the rate of 40 KMPH, and on the 3rd day for 15 hrs @ 32 KMPH. Compute HM and weighted mean and compare them. Harmonic Mean x 1 x 48 0.0208 40 0.025 32 0.0312 ∑1 x = 0.0770 Data: 10 hrs @ 48 KMPH 12 hrs @ 40 KMPH 15 hrs @ 32 KMPH N HM = ∑1 x 3 = 0.0770 HM = 38.91 Weighted Mean w x wx 10 48 480 12 40 480 15 32 480 Σw = 37 Σwx = 1440 ∑ wx Weighted Mean = x = ∑w 1440 = 37 x = 38.91 Both the same HM and WM are same. 8
9. 9. 3. Find HM for the following data. 1 1 Class (CI) Frequency (f) Mid point (m) Reciprocal   f  m m 0 – 10 5 5 0.2 1 10 – 20 15 15 0.0666 0.999 20 – 30 25 25 0.04 1 30 – 40 8 35 0.0285 0.228 40 – 50 7 45 0.0222 0.1554 Σf = 60 1 Σ f   = 3.3824 m N HM = 1 ∑f  m 60 = 3.3824 HM = 17.73 Relationship between Mean, Geometric Mean and Harmonic Mean. 1. If all the items in a variable are the same, the arithmetic mean, harmonic mean and Geometric mean are equal. i.e., x = GM = HM . 2. If the size vary, mean will be greater than GM and GM will be greater than HM. This is because of the property that geometric mean to give larger weight to smaller item and of the HM to give largest weight to smallest item. Hence, x > GM > HM . Median Median is the value of that item in a series which divides the array into two equal parts, one consisting of all the values less than it and other consisting of all the values more than it. Median is a positional average. The number of items below it is equal to the number. The number of items below it is equal to the number of items above it. It occupies central position. Thus, Median is defined as the mid value of the variants. If the values are arranged in ascending or descending order of their magnitude, median is the middle value of the number of variant is odd and average of two middle values if the number of variants is even. Ex: If 9 students are stand in the order of their heights; the 5th student from either side shall be the one whose height will be Median height of the students group. Thus, median of group is given by an equation. 9
10. 10.  N + 1 Median =    2  Ex 1. Find the median for following data. 22 20 25 31 26 24 23 Arrange the given data in array form (either in ascending or descending order). 20 22 23 24 25 26 31  N + 1 th  7 + 1 8 th Median is given by   item =  2  = 4 Median = 4 item.  2    2. Find median for following data. 20 21 22 24 28 32  N + 1 th  6 + 1 th Median is given by   item =  2  Median = 3.5 item.  2    The item lies between 3rd and 4. So, there are two values 22 and 24. The median value will be the mean values of these two values.  22 + 24  Median =   = 23  2  Discrete Series – Median In discrete series, the values are (already) in the form of array and the frequencies are recorded against each value. However, to determine the size of  N + 1 th median   item, a separate column is to be prepared for cumulative  2  frequencies. The median size is first located with reference to the cumulative frequency which covers the size first. Then, against that cumulative frequency, the value will be located as median. 10
11. 11. Ex: Find the median for the students’ marks. Obtained in statistics No. of Cumulative Marks (x) students (f) frequency 10 5 5 20 5 10 30 3 13 Just above 34 40 15 28 is 58. Against 58 c.f. the 50 30 58 value is 50 60 10 68 which is median value N = 68 Ex: In a class 15 students, 5 students were failed in a test. The marks of 10 students who have passed were 9, 6, 7, 8, 9, 6, 5, 4, 7, 8. Find the Median marks of 15 students. Marks No. of students (f) cf 0 1 2 3 5 4 1 6 5 1 7 6 2 9 7 2 11 8 2 13 9 2 15 Σf = 15 N + 1th Median = item 2 15 + 1 Me = = 8th 2 Me 8th item covers in cf of 9. the marks against cf 9 is 6 and hence Median = 6 11
12. 12. Continuous Series The procedure is different to get median in continuous series. The class intervals are already in the form of array and the frequency are recorded against each th n class interval. For determining the size, we should take item and median class 2 located accordingly with reference to the cumulative frequency, which covers the size first. When the median class is located, the median value is to be interpolated using formula given below. h N  Median =  +  2 − C f    0 + 1 Where  = where,  0 is left end point of N/2 class and l1is right end 2 point of previous class. h = Class width, f = frequency of median clas C = Cumulative frequency of class preceding the median class. Ex: Find the median for following data. The class marks obtained by 50 students are as follows. Cum. CI Frequency (f) frequency (cf) 10 – 15 6 6 15 – 20 18 24 20 – 25 9 33  N/2 class 25 – 30 10 43 30 – 35 4 47 35 – 40 3 50 Σf = N = 50 N 50 = = 25 2 2 Cum. frequency just above 25 is 33 and hence, 20 – 25 is median class.  0 + 1 = 2 20 + 20 = 20 2  = 20 h = 20 – 15 = 5 12
13. 13. f=9 c = 24 h N  Median =  +  2 − C f   5 Median = 20 + [ 25 − 24] 9 5 = 20 + 9 Median = 20.555 Ex: Find the median for following data. Mid values (m) 115 125 135 145 155 165 175 185 195 Frequencies (f) 6 25 48 72 116 60 38 22 3 The interval of mid-values of CI and magnitudes of class intervals are same i.e. 10. So, half of 10 is deducted from and added to mid-values will give us the lower and upper limits. Thus, classes are. 115 – 5 = 110 (lower limit) 115 – 5 = 120 (upper limit) similarly for all mid values we can get CI. Cum. CI Frequency (f) frequency (cf) 110 – 120 6 6 120 – 130 25 31 130 – 140 48 79 140 – 150 72 151 150 – 160 116 267 N/2 class 160 – 170 60 327 170 – 180 38 365 180 – 190 22 387 190 – 200 3 390 Σf = N = 390 N 390 = 2 2 = 195 Cum. frequency just above 195 is 267. 13
14. 14. Median class = 150 – 160 150 + 150  = = 150 2 h = 116 N/2 = 195 C = 151 h = 10 h N  Median =  +  2 − C f   10 Median = 150 + [195 −151] 116 Median = 153.8 Merits of Median a. It is simple, easy to compute and understand. b. It’s value is not affected by extreme variables. c. It is capable for further algebraic treatment. d. It can be determined by inspection for arrayed data. e. It can be found graphically also. f. It indicates the value of middle item. Demerits of Median a. It may not be representative value as it ignores extreme values. b. It can’t be determined precisely when its size falls between the two values. c. It is not useful in cases where large weights are to be given to extreme values. 14