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*10^(-6) \\ (0.992-0.995)^2=(-0.003)^2=0.000009=9*10^(-6) \\ (0.998-0.995)^2=(0.003)^2=0.000009=9*10^(-6) \end{gathered}`

And their sum goes into the equation:

`\begin{gathered} \sum ^{}_i(x_i-\bar{x})^2=1*10^(-6)+9*10^(-6)+9*10^(-6)=1.9*10^(-5) \\ s=\sqrt[]{\frac{\sum^{}_i(x_i-\bar{x})^2}{n-1}}=\sqrt[]{(1.9*10^(-5))/(3-1)}=\sqrt[]{(1.9*10^(-5))/(2)}=\sqrt[]{9.5*10^(-6)}\approx0.003 \end{gathered}`

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