A student was interested in the cost of renting an apartment close to her college. A sample of advertised rents found the following data (in dollars per month): 200 200 360 345 230 300 275 250 320 365 455 290 The standard deviation of these rents is
Solution:
x | dx = x - A = x - 276 | dx2 |
200 | -76 | 5776 |
200 | -76 | 5776 |
360 | 84 | 7076 |
345 | 69 | 4761 |
230 | -46 | 2116 |
300 | 24 | 576 |
275 | -1 | 1 |
250 | -26 | 676 |
320 | 44 | 1936 |
365 | 89 | 7921 |
455 | 179 | 32041 |
290 | 14 | 196 |
x = 3590 | dx = 2 | dx2 = 145008 |
Sample standard deviation is S
S = ( dx2 - (( dx )2 / n) n -1 )
= (145008 - (( 2 ) 2 / 13 ) 12 )
= ( 145008 - 0.3077 / 12 )
= (145007.6923 /12 )
= 12083.9744
= 109.9271
The standard deviation of these rents is 110
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