Answer
A) A: $2.4\ g/cm^3$ B: $2.703\ g/cm^3$
B) A: $11.18\%$ B: $0.04\%$
C) A: $0.216\ g/cm^3$ B: $0.0016\ g/cm^3$
D) Method B is more precise and more accurate.
Work Step by Step
A) Results that are very different from the others (outliers, like 5.811 in method B) should not be included in the calculations.
Averages: A: $(2.2+2.3+2.7+2.4)/4=2.4\ g/cm^3$
$\quad\quad\quad\quad$B: $(2.703+2.701+2.705)/3=2.703\ g/cm^3$
B) A: $E=|2.4-2.702|/2.702\times100\%=11.18\%$
$\quad$B: $E=|2.703-2.702|/2.702\times 100\%=0.04\%$
C) A: $\sigma=\sqrt{\frac{(-0.2)^2+(-0.1)^2+(0.3)^2+(0)^2}3}=0.216\ g/cm^3$
$\quad$B: $\sigma=\sqrt{\frac{(0)^2+(-0.001)^2+(0.002)^2}2}=0.0016\ g/cm^3$
D) Method B is more precise, because it had a smaller standard deviation, and it is more accurate because it had a smaller experimental error.
