Error of an area

A rectangular parcel of land was surveyed. The measurement for side X

was 339.21 ft with an error of ±0.05 ft. Side Y was measured as 563.67 ft,

with an error of ±0.09 ft. What is the area of the parcel and what is the

expected error in the area?

Answer: Area = 191,202 ft2 or 4.389 ac.; standard error = ±41.5 ft2

Does anyone know how the standard error was calculated ?

Tina - let me know if you want a visually better response, it's hard to type in equations here.

SE = sigma / sqrt i.e. the standard deviation / the square of the number of measurements.

SE = 58.7 / sqrt(2) = 41.5 ft^2

This problem is kind of tricky, but it really is a simple problem if you remember the definition of the standard error and sigma:

sigma = (1 / N) * sum(sqrt((Xi - Xbar)^2))

sigma = (1 / 1) * sqrt( [(339.21 +.05)*(563.67 + 0.09) - 339.21*563.67]^2 ) = 58.7 ft^2

This is a good example of problems that can trip you up if you try to make things to complicated; relax, look at the equations, understand what they mean, then plug and chug.

Good luck.

Tina:

To find the standard error of the area of a rectangle, you use the formula A=X*Y

X = 339.21 +/- 0.05

Y = 563.67 +- 0.09

stnderrorArea = SQRT[Y^2 * (0.05)^2 + X^2 * (0.09)^2]

therefore the answer comes out to 41.5 sf

Another way would be (339.21+0.05) * (563.67+0.09) = 191262 sf

Difference = 191262 - 191203 = 59

Stnd error = 59/sqrt(2) = 41.7 sf

Excellent answer, single.

Easy to understand.....

That will help more than just Tina....

ktulu

Thanks guys. Understood !!!

@bridge_guy,

41.7 is actually listed as one of the answers but is incorrect. The correct answer is 41.5. So I guess singlespeed is right.

^ Yea I rounded up for simplicity purposes but its Surveying (duh) where precision counts. It does work, if you carry out the numbers, but I guess 41.7 sf was to trip up anyone who rounded. Thanks for the heads up though

I reread my reply and one thing I forgot to mention was that sigma was calculated with a sample size of 1 (N = 1 for the area measurement) while the standard error was calculated with a sample size of 2 (n = 2, because of length x width). The tricky part to the problem is realizing what the proper sample sizes are.

It really is a simple problem that forces you to think about it for a minute; when you're feel like you're pressed for time and anxious, this type of problem can trip you up.

I can't stress enough that there is no room for panic on this test. I went through my exam book three times in the am (1st - problems I knew, 2nd - problems that I had to look something up, and 3rd - narrowed my choices and best guess). In the pm, I probably went through that test book 5 or 6 times.

If you have prepared enough, you'll have plenty of time if you don't panic and waste time. If you can't answer the problem in 2 minutes, skip it and come back to it. Maybe the next time (or the 3rd of 4th time) you read the problem, you'll recognize how to solve it in 2 minutes. It worked for me.

Good luck all!