The ratio of the number of atoms of the isotopes ${}^{\mathbf{69}}\mathbf{Ga}$and${}^{\mathbf{71}}\mathbf{Ga}$in eight samples from different sources was measured in an effort to understand differences in reported values of the atomic mass of gallium:

Find the (a) mean, (b) standard deviation, (c) variance, and (d) standard deviation of the mean. (e) Write the mean and standard deviation together with an appropriate number of significant digits.

• The standard deviation of a bunch of numbers is the distance between them and the mean.
• The variance is a measure of how much each point deviates from the mean on average.

(a)

The mean,

$$\begin{gathered} \overline{x}=\frac{{\displaystyle \sum _i}{x}_i}{n} \\ =\frac{1.527660+1.52974+.....+1.52685+1.52793}{8} \\ =1.52766625 \\ =1.52767 \end{gathered}$$

(b)

The standard deviation,

$$\begin{gathered} s=\sqrt{\frac{{\displaystyle \sum _i}{\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(1.52660-1.527666\right)}^2+{\left(1.52974-1.527666\right)}^2+.....+{\left(1.52793-1.527666\right)}^2}{7}} \\ =1.259909\times {10}^{-3} \\ =1.26\times {10}^{-3} \end{gathered}$$

(c)

For the variance simply square the computed value for standard deviation:

$$\begin{gathered} s^2={\left(1.26\times {10}^{-3}\right)}^2 \\ =1.59\times {10}^{-6} \end{gathered}$$

(d)

The standard deviation of the mean of sets of values,

$$\begin{gathered} {\sigma }_n=\frac{\sigma }{\sqrt{n}} \\ {\sigma }_8=\frac{\sigma }{\sqrt{8}} \\ =\frac{1.26\times {10}^{-3}}{\sqrt{8}} \\ =4.454772721\times {10}^{-4} \\ =4.5\times {10}^{-4} \end{gathered}$$

(e)

The sample mean and standard deviation calculated values.

Then, considering the number of significant figures write them in the form of $\overline{x}\pm s$Then, via substitution, $1.{527}_7\pm 0.{001}_3$.