You and your lab partner are asked to determine the density of an aluminum bar.

Question

You and your lab partner are asked to determine the density of an aluminum bar. The mass is known accurately (to four significant figures). You use a simple metric ruler to measure its dimensions and obtain the results for Method A. Your partner uses a precision micrometer and obtains the results for Method B. Method A (g/cm3) Method B (g/cm3) 2.4 2.699 2.5 2.697 2.9 2.701 2.6 5.803 The accepted density of aluminum is 2.702 g/cm3. Calculate the standard deviation for each set of data. The average value for Method A is 2.6 g/cm3 and the average value for Method B is 2.699 g/cm3. Method A Method B

Explanation / Answer

Variance = (sum of (x - average)^2)/n

Method A:

Variance = [(2.4 - 2.6)^2 + (2.5 - 2.6)^2 + (2.9 - 2.6)^2 + (2.6 - 2.6)^2]/4 = 7/200

So standard deviation for A = sqrt(Variance) = sqrt(7/200) = 0.1871 g/cm^3

Method B:

Variance = [(2.699 - 2.699)^2 + (2.697 - 2.699)^2 + (2.701 - 2.699)^2 + (5.803 - 2.699)^2]/4 = 2.4087

So standard deviation for B = sqrt(Variance) = sqrt(2.4087) = 1.552 g/cm^3

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