# Homework Help: Error percentage

1. Dec 9, 2011

### sergey90

1. The problem statement, all variables and given/known data
Estimate the percent error in measuring a distance of about 75cm with a meter stick.

2. Relevant equations

3. The attempt at a solution
I'm assuming one could accurately measure this value to a couple mm. so it would be 75+-0.2cm. Now the book gives an answer of 0.13% i just have no clue how they get there. Thanks in advance.

2. Dec 9, 2011

### Delphi51

If the meter stick has mm's as the smallest division, you would be rounding at most .5 mm on each end. Divide by 75 and multiply by 100 to convert to per cent.

3. Dec 10, 2011

### PeterO

Your uncertainty of 0.2 cm means 0.2 in 75, which is 2 in 750 which is 1 in 375

1/375 * 100 = 0.26666 %

If you had thought you could get it to within 1 mm, the uncertainly would have been 0.13 %

4. Dec 10, 2011

### technician

I would say the error on any measuring instrument is +/- 1 scale division or +/1 1 in the last digit of a digital meter.
If the metre stick has mm divisions then I would say the error is +/- 1mm which is +/-1 in 750 (similar to PeterO with his +/-)
The question refers to a distance of ABOUT 75cm so it seems to me they want a rough estimate of the error rather than an answer to several significant figures.

5. Dec 14, 2011

### sergey90

thanks for the help guys. One last clarification on the matter: in previous problems of fractional errors if for example we have a value of 9.69+-0.07cm^2 the book just divides 0.07 by 9.69 and says thats my fractional error % without multiplying by 100. Why so?

6. Dec 14, 2011

### PeterO

Perhaps it was given as a fractional error, to convert a fraction to a percentage, you multiply by 100.

It is unusual to fractional error % - usually either fractional error or percentage error.