Question

A sample of college students was asked how much they spent monthly on pizza. Approximate the standard deviation for the cost.

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Answer #1
Class interval Midpoint, x Frequency, f fx fx²
10 - 19.99 14.995 10 149.95 2248.50025
20 - 29.99 24.995 18 449.91 11245.50045
30 - 39.99 34.995 26 909.87 31840.90065
40 - 49.99 44.995 11 494.945 22270.05028
50 - 59.99 54.995 8 439.96 24195.6002
Total 73 2444.635 91800.5518

Standard deviation, s = √((∑fx² - (∑fx)²/n)/(n-1))

= √((91800.5518 - 2444.635²/73)/(73- 1)) = 11.75

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