A distance was measured by pacing as 267 ft. with a standard deviation of ±3 ft. It was then measured as XXXXXXXXXXft. with a steel tape and had a standard deviation of +0.05ft. Finally, it was...

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A distance was measured by pacing as 267 ft. with a standard deviation of ±3 ft. It was then measured as 269.08 ft. with a steel tape and had a standard deviation of +0.05ft. Finally, it was measured as 268.99 ft. with an EDM instrument. The EDM instrument and reflector setup standard deviations were ±0.005 ft and ±0.01ft. respectively, and the manufacturer's estimated standard deviation for the EDM instrument is ±( 5 mm + 5 ppm). What are the most probable value for the distance and its standard deviation? What is the computed standard deviation for each weighted observation?

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Answered Same DayDec 22, 2021

Answer To: A distance was measured by pacing as 267 ft. with a standard deviation of ±3 ft. It was then...

David answered on Dec 22 2021

120 Votes

Solution
Distance when it was measured by pacing = 267 ft ± 3 ft
Distance when it was measured by steel tape = 269.08 ft ± 0.05 ft
Distance when it was measured by EDM instruments = 268.99 ft
The EDM instrument and reflector setup having standard deviation = ± 0.005 ft and ± 0.01 ft
The net impact of SD = sqrt (0.005^2 + 0.01^2) = 0.0112 ft
Manufacturer estimated standard deviation for EDM instruments- = ±(5 mm + 5 ppm)
= ±(0.017 ft + 5 ppm)
For the measurement of 268.99 ft =>...

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A distance was measured by pacing as 267 ft. with a standard deviation of ±3 ft. It was then measured as XXXXXXXXXXft. with a steel tape and had a standard deviation of +0.05ft. Finally, it was...

Answer To: A distance was measured by pacing as 267 ft. with a standard deviation of ±3 ft. It was then...

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