Measures of Position and dispersion.pdf

MEASURES OF POSITION (Partition Values) The values which divide the data points into a number of equal parts are called partition values. The commonly used partition values are 1. Median- divide data points into two equal parts 2. Quartiles- divide data points into four equal parts 3. Deciles- divi...

MEASURES OF POSITION (Partition Values) The values which divide the data points into a number of equal parts are called partition values. The commonly used partition values are 1. Median- divide data points into two equal parts 2. Quartiles- divide data points into four equal parts 3. Deciles- divide data points into ten equal parts 4. Percentiles- divide data points into hundred equal parts Quartiles ℎ 𝑟𝑁 𝑄𝑟 = 𝑙 + −𝑐 𝑓 4 Deciles ℎ 𝑟𝑁 𝐷𝑟 = 𝑙 + −𝑐 𝑓 10 Percentiles ℎ 𝑟𝑁 𝑃𝑟 = 𝑙 + −𝑐 𝑓 100 MEASURES OF DISPERSION In a frequency distribution, though the values cluster around an average, most of them differ from it. This property of deviation of values from the average is called variation or dispersion. The various measures of dispersion are 1. Range 2. Quartile deviation 3. Mean deviation 4. Standard deviation  Range (R) Range is the difference between the highest (H) and lowest (L) values in the data. 𝑅 =𝐻−𝐿 When we have to compare variability of two series which differ in their averages or are of different units, we calculate coefficient of dispersion which are independent of the units of measurement. 𝐻−𝐿 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑅 = 𝐻+𝐿 Since 𝑅 is based on extreme values, it is not a reliable measure of dispersion.  Quartile deviation (Q) 1 𝑄= 𝑄3 − 𝑄1 2 𝑄3 − 𝑄1 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑄 = 𝑄3 + 𝑄1 where 𝑄1 and 𝑄3 are the first and third quartiles respectively. Quartile deviation considers only 50% of the data but ignores the other 50%. Hence it is not a reliable measure of dispersion.  Mean deviation Mean deviation from a central value 𝐴 is 𝑛 1 𝑀. 𝐷 𝐴 = ෍ 𝑓𝑖 |𝑥𝑖 − 𝐴| 𝑁 𝑖=1 where 𝐴 can be mean, median or mode. 𝑀. 𝐷(𝐴) 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑀. 𝐷 𝐴 = 𝐴 It is based on all observations. Hence it is better than range and quartile deviation but |𝑥𝑖 − 𝐴| ignores the signs of the devaitions. NOTE : Mean deviation is least when taken from the median. Problem 1. Calculate : a. Quartile Deviation and b. Mean Deviation from mean for the following data: MARKS 0-10 10-20 20-30 30-40 40-50 50-60 60-70 No. of 6 5 8 15 7 6 3 Students Ans. a. 11.23, b. 13.184  Standard deviation (𝝈) 𝑛 1 2 𝜎= ෍ 𝑓𝑖 𝑥𝑖 − 𝑥ҧ 𝑁 𝑖=1 𝜎 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑥ҧ Squaring 𝑥𝑖 − 𝑥ҧ overcomes the drawback of ignoring the signs in mean deviation. Standard deviation is the best and most powerful measure of dispersion. 1 Variance, 𝜎2 = σ𝑛𝑖=1 𝑓𝑖 𝑥𝑖 − 𝑥ҧ 2 𝑁 𝜎 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑡𝑖𝑜𝑛 = 100 × 𝑥ҧ NOTE: For comparing variability of two series, we calculate the coefficient of variation for each series. The series having lesser value of this coefficient is more consistent than the other. On simplifying the formula for 𝜎𝑥2 , we get 𝑛 𝑛 2 1 2 1 𝜎𝑥2 = ෍ 𝑓𝑖 𝑥𝑖 − ෍ 𝑓𝑖 𝑥𝑖 𝑁 𝑁 𝑖=1 𝑖=1 Short-cut method (For large values of x and f) 𝑥𝑖 −𝐴 Let 𝑑𝑖 = , then 𝜎𝑥2 = ℎ2 𝜎𝑑2 ℎ If 𝑛1 and 𝑛2 are the sizes, 𝑥1 and 𝑥2 are the means and 𝜎1 and 𝜎2 are the standard deviations of two series respectively, then the standard deviation 𝜎 of the combined series of size 𝑛1 + 𝑛2 is 2 2 2 2 2 𝑛 𝜎 1 1 + 𝑑1 + 𝑛 𝜎 2 2 + 𝑑2 𝜎 = 𝑛1 + 𝑛2 𝑛1 𝑥1 +𝑛2 𝑥2 where 𝑑1 = 𝑥1 − 𝑥,ҧ 𝑑2 = 𝑥2 − 𝑥ҧ and 𝑥ҧ = 𝑛1 +𝑛2 NOTE: For any set of values, 𝜎 ≥ 0.  Problems 11. Calculate mean deviation from 𝑥ҧ for the following data : 140, 147, 143, 146, 144 Ans. 2 12. Compute mean deviation from median for the following Marks 0-10 10-20 20-30 30-40 40-50 50-60 No. of students 2 10 20 15 10 3 Ans. 9.93 13. Calculate standard deviation for the following table giving the age distribution of 542 members Age (years) 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No. of persons 3 61 132 153 140 51 2 14. For a group of 200 candidates, the mean and standard deviation of scores were found to be 40 and 15 respectively. Later on, it was discovered that the scores 43 and 35 were misread as 34 and 53 respectively. Find the corrected mean and standard deviation corresponding to the corrected figures. 15. An analysis of monthly wages paid to the workers of two firms A and B belonging to the same industry gives the following results Firm A Firm B No. of workers 500 600 Average monthly wage Rs. 186.00 Rs. 175.00 Variance of distribution of 81 100 wages i. Which firm, A or B, has a larger wage bill? ii. In which firm, A or B, is there greater variability in individual wages? iii. Calculate (a) average monthly wage and (b) the variance of the distribution of wages of all the workers in the firms A and B taken together.