SOLUTION: Find the range, standard deviation, and variance of A survey of 10 fast-food restaurant noted the number of calories in a mid-sized hamburger. The following are results 514 507

Algebra ->  Finance -> SOLUTION: Find the range, standard deviation, and variance of A survey of 10 fast-food restaurant noted the number of calories in a mid-sized hamburger. The following are results 514 507      Log On


   

Question 1158372: Find the range, standard deviation, and variance of
A survey of 10 fast-food restaurant noted the number of calories in a mid-sized hamburger. The following are results
514 507 502 498 496 506 458 478 463 514
Answer by KMST(5396) About Me  (Show Source):
You can put this solution on YOUR website!
Range, standard deviation, and variance are measures of the dispersion of the data in a set.
A long time ago, in college, I would calculate them using paper and pen, based on their definitions.
Nowadays, a calculator, or computer software can calculate them for us easily, but we have to understand what this measure tell us and exactly what our devices.

Range:
The range is the difference between the smallest and greatest values.
Range=514-458=highlight%2856%29

Variance:
The mean (average) of all 10 values is
%28514%2B507%2B502%2B498%2B496%2B506%2B458%2B478%2B463%2B514%29%2F10=4936%2F10=493.6
The deviations from the mean for the individual values are:
514-493.6=20.4 (514 appears twice)
507-493.6=13.4
502-493.6=8.4
498-493.6=4.4
496-493.6=2.4
506-493.6=12.4
458-493.6=-35.6%7D%7D%0D%0A%7B%7B%7B478-493.6=-15.6
463-493.6=-30.6
Their squares are:
20.4%5E2=416.16 (514 appears twice)
13.4%5E2=179.56
8.4%5E2=70.56
4.4%5E2=19.36
2.4%5E2=5.76
12.4%5E2=153.76
-35.6%5E2=1267.36%7D%7D%0D%0A%7B%7B%7B-15.6%5E2=243.36
-30.6%5E2=936.36
The variance is the average of those squares:


Standard deviation:
The standard deviation, as asked in this question, is the square root of the variance as calculated above:
sqrt%28370.84%29=highlight%2819.26%29 (rounded)
Software that calculates statistics would have to forms of standard deviation:
the standard deviation of a population, calculated as above, and
the standard deviation of a sample, used to estimate the dispersion of the whole population of data based on a sample of N items. In the second case, the sum of squares in divided by N-1 instead.