91Ó°ÊÓ

• The standard deviation is a measure of how far something deviates from the mean (for example, spread, dispersion, or spread). A "typical" variation from the mean is represented by the standard deviation.
• Because it returns to the data set's original units of measurement, it's a common measure of variability.
• The standard deviation, defined as the square root of the variance, is a statistic that represents the dispersion of a dataset relative to its mean.

a)

Calculate the mass of sample with the following equation

\(\begin{aligned}{}m{R^2} &= {K_S}\\m &= \frac{{{K_S}}}{{{R^2}}}\\m &= \frac{{20\;{\rm{g}}}}{{{2^2}}}\\m &= 5\;{\rm{g}}\end{aligned}\)

Hence, the mass of the sample for the given standard deviation is \(5\;{\rm{g}}\).

b)

Use the equation 28-7 with \({s_s} = 0.02\)and \(e = 0.015\), the initial value of \(t\) is found on the Pg.72 Table \(4 - 4\) for confidence \(90\% \) :

\(\begin{aligned}{}n &= \frac{{{t^2}s_s^2}}{{{e^2}}}\\n &= \frac{{{{1.645}^2} \times {{0.02}^2}}}{{{{0.015}^2}}}\\n& = 4.8\end{aligned}\)

Next, for \(n = 5\) there are \({4^\circ }\) of freedom so \(t = 2.132\) which gives the following value of \(n\) :

\(n = \frac{{{t^2}s_s^2}}{{{e^2}}}\)

\(\begin{aligned}{}n &= \frac{{{{2.132}^2} \times {{0.02}^2}}}{{{{0.015}^2}}}\\n &= 8.1\end{aligned}\)

Then if we use in the same equation the \({7^\circ }\) of freedom \(t = 1.895\) this would give us the following value of \(n\) :

\(\begin{aligned}{}n &= \frac{{{t^2}s_s^2}}{{{e^2}}}\\n &= \frac{{{{1.895}^2} \times {{0.02}^2}}}{{{{0.015}^2}}}\\n &= 6.4\end{aligned}\)

Then with \(t = 2.015 \to n = 7.2\) and \(t = 1.943 \to n = 6.7\)So, we would use a total of \(7\) samples.

Therefore, in this case we should use \(7\) samples,\(n = 4.8,8.1,6.4,7.2,6.7\)