91影视

The following sample observations are provided for young鈥檚 modulus in units of GPa.

a.

Let x represents the sample values.

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{116.4 + 115.9 + 114.6 + 115.2 + 115.8}}{5}\\ &=& \frac{{577.9}}{5}\\ \approx 115.58\end{array}\)

Thus, the sample mean is 115.58GPa.

The deviations are computed by subtracting the mean value from each observation.

Mathematically,\({{\bf{d}}_{\bf{i}}}{\bf{ = }}\left( {{{\bf{x}}_{\bf{i}}}{\bf{ - \bar x}}} \right)\)

The table representing the deviations is as follows,

b.

The sample variance is given as,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{\sum {{d_i}^2} }}{{n - 1}}\\ &=& \frac{{{S_{xx}}}}{{n - 1}}\end{array}\)

The calculations are represented as,

Substituting the values in the formula, we have,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{1.928}}{{5 - 1}}\\ &=& 0.482\end{array}\)

Therefore, the sample variance for the provided data is 0.482GPa square.

The sample standard deviation is given as,

\(\begin{array}{c}s &=& \sqrt {{s^2}} \\ &=& \sqrt {0.482} \\ &=& 0.694\end{array}\)

Therefore, the sample standard deviation for the provided data is 0.694GPa.

c.

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 calculations to compute the variance are as follows,

Substituting the values,

\(\begin{array}{c}{s^2} &=& \frac{{{S_{xx}}}}{{n - 1}}\\ &=& \frac{{66795.61 - \frac{{{{\left( {577.9} \right)}^2}}}{5}}}{{5 - 1}}\\ &=& \frac{{1.928}}{4}\\ &=& 0.482\end{array}\)

Therefore, the sample variance is 0.482GPa square.

d. Subtracting 100 from each observation to obtain transformed values.

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{16.4 + 15.9 + 14.6 + 15.2 + 15.8}}{5}\\ \approx 15.58\end{array}\)

Thus, the sample mean is 15.58.

The sample variance is given as,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{{S_{xx}}}}{{n - 1}}\end{array}\)

The calculations are as follows,

Substituting the values, we have,

The sample variance is,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{1.928}}{{5 - 1}}\\ &=& 0.482\end{array}\)

Therefore, the sample variance is 0.482.

It can be observed that the sample variance of the transformed data is the same as the sample variance of the original data. This implies that the variance is independent of the change of origin.