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The data is provided as,

The size of the sample is 17.

a.

Let x represents the sample values.

The symbol \(\sum {} \)shows the summation of a range of values.

The table representing the calculations are as follows,

Therefore, the values are,

\(\begin{array}{c}\sum {{x_i}} &=& 2.75 + 2.62 + ... + 3.01\\ &=& 56.8\\\sum {x_i^2} &=& {2.75^2} + {2.62^2} + ... + {3.01^2}\\ &=& 197.804\end{array}\)

Therefore, \(\sum {{x_i}} = 56.8\;m{m^2}\) and \(\sum {x_i^2} = 197.804\;m{m^2}\;square\)

b.

Referring to part a, the values are,

\(\sum {{x_i}} = 56.8\)and\(\sum {x_i^2} = 197.804\)

The sample variance is given as,

\({s^2} = \frac{{{S_{xx}}}}{{n - 1}}\)

Where,

\({S_{xx}} = \sum {x_i^2} - \frac{{{{\left( {\sum {{x_i}} } \right)}^2}}}{n}\)

The sample variance is computed as,

\(\begin{array}{c}{s^2} &=& \frac{{{S_{xx}}}}{{n - 1}}\\ &=& \frac{{197.804 - \frac{{{{\left( {56.8} \right)}^2}}}{{17}}}}{{17 - 1}}\\ &=& 0.5016\end{array}\)

Therefore, the sample variance is 0.5016\(m{m^2}\;square\).

The sample standard deviation is computed as,

\(\begin{array}{c}s &=& \sqrt {{s^2}} \\ &=& \sqrt {0.5016} \\ &=& 0.708\end{array}\)

Therefore, the sample standard deviation is 0.708\(m{m^2}\).