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Mth115 Measure of Dispersion (4.3) Ten students from the Northern Washington University were asked how many minutes they studied each night. Also, ten students from Southern Washington University were asked the same question. Here’s the results: [A] Let’s find the Standard Deviation for the NWU data! You’ll need the mean, which is ______ Minutes Deviation (x) (x – x ) 50 60 75 120 80 95 120 110 80 80 (Deviation)2 (x – x )2 i) After finishing the table, we have (x - x )2 = ___________ ii) The Variance is s2 = (x - x )2 n–1 = ___________ iii) The Standard Deviation is s = s2 = ___________ Note: The Standard Deviation is the square root of the Variance. The larger the Standard Deviation, the more spread out the data is from the mean. There’s another way of finding Variance, which we’ll try on the next page. [B] What percentage of the NWU data is … …within 1 Standard Deviation of the Mean: [ x – 1s, x + 1s] ? ___________ …within 2 Standard Deviation of the Mean: [ x – 2s, x + 2s] ? ___________ 1 [C] Let’s find the Standard Deviation for the SWU data! Minutes (x) 75 80 80 85 85 85 90 90 95 105 (Minutes 2) (x 2) (i) After finishing the table, you’ll need (x) __________ (ii) Now find [ (x) ]2 = __________ (iii) You want [ (x) ]2 n = __________ (iv) Also, you’ll need (x2 ) = __________ (v) And now the Variance s2 = (iv) – (iii) n–1 = 2 [(x )] (x2 ) n __________________ n–1 n = __________ = __________ (vi) The Standard Deviation is s = s2 [D] What percentage of the SWU data is … …within 1 Standard Deviation of the Mean: [ x – 1s, x + 1s] ? __________ …within 2 Standard Deviation of the Mean: [ x – 2s, x + 2s] ? __________ …within 3 Standard Deviation of the Mean: [ x – 3s, x + 3s] ? __________ 2 [C] We can use a similar technique as we did in [B] to find the Standard Deviation for Grouped Data. Suppose the following table shows data for the ages of people who earned the minimum wage. Let’s find the Standard Deviation for this data. y = Age f = Frequency (In Thousands) 15 y < 19 514 19 y < 25 570 25 y < 30 411 35 y < 40 309 40 y < 50 225 50 y < 70 108 x = Midpoint (Of Age Interval) fx DON’T FORGET n = f = __________ (i) After finishing the table, you’ll need ( f x) = __________ (ii) Now find [ ( f x) ]2 = __________ (iii) You want [ ( f x) ]2 n = __________ (iv) Also, you’ll need ( f x2 ) = __________ f x2 (v) And now the Variance s2 = (iv) – (iii) n–1 = ( f x2 ) [(f x )] n __________________ n–1 n 2 = __________ = __________ (vi) The Standard Deviation is s = s2 3