An intern measures the dimensions of a rectangular excavatio

An intern measures the dimensions of a rectangular excavation to be: Length- 302+/-0.79 feet width = 98 +/-0.9 feet Depth = 19 +/-1.2 feet What is the most probable value for the excavation volume in cubic yards (also written as yd3 or CY)? Round your answer to the nearest cubic yard.

Solution

to calculate most possible to nearest cubic yard

there can be 2*2*2 = 8 combination

that is

(302+0.79)*(98+0.9)(19+1.2) ft³ = 604907.8062 ft³ = 22403.992822222 yd³

(302+0.79)*(98+0.9)(19-1.2) ft³=533037.5718 ft³ = 19742.1322888889 yd³

(302+0.79)*(98-0.9)(19+1.2) ft³=593898.3618 ft³ = 21996.2356222222 yd³

(302+0.79)*(98-0.9)(19-1.2) ft³=523336.1802 ft³ = 19382.821488889 yd³

(302-0.79)*(98+0.9)(19+1.2) ft³ = 601751.3138 ft³ = 22287.085696296 yd³

(302-0.79)*(98+0.9)(19-1.2) ft³=530256.1082ft³ = 19639.115118519 yd³

(302-0.79)*(98-0.9)(19+1.2) ft³=590799.3182ft³ = 21881.45622963 yd³

(302-0.79)*(98-0.9)(19-1.2) ft³=520605.3398ft³ = 19281.679251852 yard ³

max value = 22403.992822222 yard ³

min value =  19281.679251852 yard ³

consider linear variation

most provable value = (max + min )/2 =20842.836 = 20843 yd³