Question

How do I do the last two?the actual mean is 0.995

Answer 1

The standard deviation here uses the sample standard deviation formula:

$s=\sqrt[]{\frac{\sum^{}_i(x_i-\bar{x})^2}{n-1}}$

So, first we make the difference between the values and the mean:

$\begin{gathered} 0.994-0.995=-0.001 \\ 0.992-0.995=-0.003 \\ 0.998-0.995=0.003 \end{gathered}$

Now, we square them:

$\begin{gathered} (0.994-0.995)^2=(-0.001)^2=0.000001=1\times10^{-6} \\ (0.992-0.995)^2=(-0.003)^2=0.000009=9\times10^{-6} \\ (0.998-0.995)^2=(0.003)^2=0.000009=9\times10^{-6} \end{gathered}$

And their sum goes into the equation:

$\begin{gathered} \sum ^{}_i(x_i-\bar{x})^2=1\times10^{-6}+9\times10^{-6}+9\times10^{-6}=1.9\times10^{-5} \\ s=\sqrt[]{\frac{\sum^{}_i(x_i-\bar{x})^2}{n-1}}=\sqrt[]{\frac{1.9\times10^{-5}}{3-1}}=\sqrt[]{\frac{1.9\times10^{-5}}{2}}=\sqrt[]{9.5\times10^{-6}}\approx0.003 \end{gathered}$

Thus, the standard deviation is approximately 0.003 g/cm³.