If the professor holds the true value of #121# #"m/s"# with #3.68%# error, then which of the following students' results is accurate enough with respect to the professor?

#a)# Erin: #"124 m/s"#
#b)# Jean-Claude: #"116 m/s"#
#c)# Suresh: #"128 m/s"#

1 Answer
Jun 9, 2016

Looks like Erin is more accurate than the professor.


Accuracy is just the closeness to the accepted value.

Precision is based on how well the value is replicated after multiple trials, and is not the same thing.

Since the professor states his/her percent error of #3.68%#, we should determine what that is.

#3.68% = 0.0368#

#0.0368xx121 = 4.45#

So, the professor's value, including the error allowed, should be within #121 pm "4.45 m/s"#. The professor's results are therefore somewhere within the range #color(blue)(116.55 - 125.45)#.

To be most accurate, you must be within that range. If you are closer to #121# than the professor, you are most accurate.

Erin: #"124 m/s"#

#121 + 4.45 = color(green)(125.45 > 124)# #color(blue)(\mathbf(sqrt""))#

So, she is closer to #121# than the professor. Therefore, she is most accurate. Specifically...

#|124 - 121|/(121)xx100%#

#= color(blue)("Percent Error" = 2.48% < 3.68%)#

Jean-Claude: #"116 m/s"#

#121 - 4.45 = color(red)(116.55 > 116)# #color(red)(\mathbf(X))#

So, he is below the right range. He is farther from #121# than the professor. Specifically...

#|116 - 121|/(121)xx100%#

#= color(red)("Percent Error" = 4.13% > 3.68%)#

Suresh: #"128 m/s"#

#121 + 4.45 = color(red)(125.45 < 128)# #color(red)(\mathbf(X))#

So, he is above the right range. He is farther from #121# than the professor. Specifically...

#|128 - 121|/(121)xx100%#

#= color(red)("Percent Error" = 5.79% > 3.68%)#