91Ó°ÊÓ

• 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,

$$\begin{gathered} \stackrel{¯}{x}=\frac{{\displaystyle \underset{i}{∑}}{x}_i}{n} \\ =\frac{1.527660+1.52974+.....+1.52685+1.52793}{8} \\ =1.52766625 \\ =1.52767 \end{gathered}$$

(b)

The standard deviation,

$$\begin{gathered} s=\sqrt{\frac{{\displaystyle \underset{i}{∑}}{\left(x_i-\stackrel{¯}{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{{\left(1.52660-1.527666\right)}^2+{\left(1.52974-1.527666\right)}^2+.....+{\left(1.52793-1.527666\right)}^2}{7}} \\ =1.259909×{10}^{-3} \\ =1.26×{10}^{-3} \end{gathered}$$

(c)

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

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

(d)

The standard deviation of the mean of sets of values,

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

(e)

The sample mean and standard deviation calculated values.

Then, considering the number of significant figures write them in the form of $\stackrel{¯}{x}Â\pm s$Then, via substitution, $1.{527}_7Â\pm 0.{001}_3$.