Measures of Dispersion Prepared by: Bhakti Joshi Date: November 30, 2011

Meaning and Importance Study of dispersion or “variation” in observations that we seek to explain Study whether observations lie around the centre or away from the centre Examples: – Wages provided in the factory – Income variations across populations – Variations in petrol prices across cities – Variations in the standard of living of people within a city/region

Tools in Measures of Dispersion Range – Difference between the highest and lowest observations Interquartile Range – Measures approximately how far from the median Variance – Measures the spread of values or observations around the mean Standard Deviation – Square root of variance

Range Difference between largest & smallest observations Ignores how data are distributed Range   x largest smallest x

Weekly WPI Indices from June 04, 2011 to November 12, ,195.4,194.9,196.7, 197.2,197.5,198.6,198.0, 199.5,197.8,199.0,200.2, 200.6,201.8,202.9,204.0, 202.4,203.4,203.8,204.5, 205.0,204.7,203.0, Step 1: Arrange all the numbers in an ascending order Step 2: Identify the minimum index Step 3: Identify the maximum index Range: 205 – = 10.1

Interquartile Range Difference between third & first quartiles (Data points are divided into four parts or by 1/4 th ) Interquartile Range = Q 3 - Q 1 Spread in middle 50% Not affected by extreme values Step 1: Determine Quartile 1, which is the data point lies across one-fourth or 25% of the observations Step 2: Determine Quartile 3, which is the data point lies across one-fourth or 25% of the observations from the median Interquartile Range: – = 5.9

Interquartile Range Quartile 1 Quartile 3 Quartile 2

Variance & Standard Deviation Most commonly used measures Consider how data are distributed Show variation about mean ( x or  ) Symbols: PopulationSample Variance Standard Deviation s  s 2  2

Population Variance Formula x 2 2     N  () xx N      ( ()) ) x  ( N Where: N - Population size x - item or observation - Population mean - sum of the values  

Population Variance Formula x 2 2     N  () xx N      ( ()) ) x  ( N Remember --- N in denominator. Use n - 1 if Sample Variance

Population Standard Deviation Formula  xixi N xxx N N            ( ((( ) )))... Square Root

Sample Variance Formula Where: n - sample size x - item or observation x - sample mean - sum of the values  n - 1 in denominator! (Use N if Population Variance) n1    S n 2 1    (x i x) 2  _ (x 1 x) 2  _  (x 2 x) 2  _ (x n x) 2  _  x)

Calculate Variance and Standard Deviation for WPI Step 1: Calculate the mean x Step 2: Determine ‘n’ Step 3: Another column determine  i.e. Step 4: Subsequent column, determine  i.e. Step 5: Calculate the sum of the column determined in Step 4 Step 6: Divide the sum with xixi x  _ x1x1 x _ x2x2 x _,,………, x 24 x  _ (x i x) 2  _ (x 1 x) 2  _ (x 2 x) 2  _,,………, (x 24 x) 2  _ n1 

Variance and Standard Deviation Mean = Sum of = Variance = / 23 = 10.1 Standard Deviation = 3.17 (x i x) 2  _

Variance & Standard Deviation for class intervals Hospital stay in days (interval) Frequency of Patients (f) Midpoint (x) fxf Mean = 1543/200= 7.715, Standard Deviation = 4.68 xixi x  _ xixi x _ 2 xixi x _ 2

Problem 1 The number of cheques cashed each day at five branches of HDFC bank during the past month had the following frequency distribution. The Director of operations knows that a standard deviation in cheque cashing of more than 200 cheques per day creates staffing and organizational problems at the branches because of the uneven workload. Should the director worry about staffing new month? ClassFrequency

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