Question

Three samples of the same metal are weighed and their masses are found to be 44.40 g, 40.58 g, and 38.35 g. The corresponding volumes are 4.8 ml, 4.7 ml, and 4.2 ml, respectively. Determine the density, d, of the metal and calculate the standard deviation of the density.

Answer 1

The density of a material is the mass of the material per unit volume. Here the weight of the same metal is 44.40g, 40.58g and 38.35g having volume 4.8 mL, 4.7 mL and 4.2 mL respectively. Thus the density of the metal as per the given data are, $\frac{44.40}{4.8}$ = 9.25g/mL, $\frac{40.58}{4.7}$ = 8.634g/mL and $\frac{38.35}{4.2}$ = 9.130g/mL respectively.

The equation of the standard deviation is √{∑(x  - $\frac{}{x}$)÷N}

Now the mean of the density is {(9.25 + 8.634 + 9.130)/3} = 9.004 g/mL.

The difference of the density of the 1st metal sample (9.25-9.004) = 0.246 g/mL. Squaring the value = 0.060.

The difference of the density of the 2nd metal sample (9.004-8.634) =0.37 g/mL. Squaring the value = 0.136.

The difference of the density of the 3rd metal sample (9.130-9.004) = 0.126 g/mL. Squaring the value 0.015.

The total value of the squared digits = (0.060 + 0.136 + 0.015) = 0.211. By dividing the digit by 3 we get, 0.070. The standard deviation will be $\sqrt{0.070}=0.265$. Thus the standard deviation of the density value is 0.265g/mL.