91影视

The data on melting point (掳C) for each of 12 specimens of the polymer using a differential scanning calorimeter.

The size of the sample is 12.

a.

The range is the difference between the largest and the smallest sample value.

The sample range is computed as,

\(\begin{array}{c}Range = 182.6 - 180.3\\ = 2.3\end{array}\)

Therefore, the sample range for the provided data is 2.3掳C.

b.

Let x represents the sample values for melting point.

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 sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{180.5 + 181.7 + 180.9 + ... + 180.5}}{{12}}\\ &=& 181.40833\\ \approx 181.41\end{array}\)

Thus, the sample mean is 181.41掳C.

The calculations that are required to compute the sample variance is as follows,

Substituting the values, the sample variance is given as,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{5.7692}}{{12 - 1}}\\ &=& \frac{{5.7692}}{{11}}\\ &=& 0.5245\end{array}\)

Thus, the sample variance for the provided data is 0.5245掳C square.

c.

Referring to the sample variance computed in part b,

The sample variance is 0.5245.

The sample standard deviation is given as,

\(\begin{array}{c}s &=& \sqrt {{s^2}} \\ &=& \sqrt {0.5245} \\ &=& 0.7242\end{array}\)

Therefore, the sample standard deviation for the provided data is 0.72442掳C.

d.

Let x represents the sample values.

The sample variance is given as,

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

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{180.5 + 181.7 + 180.9 + ... + 180.5}}{{12}}\\ &=& 181.40833\\ \approx 181.41\end{array}\)

Thus, the sample mean is 181.41.

The calculations that are required to compute the sample variance is as follows,

Substituting the values, the sample variance is given as,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {x_i^2} - \frac{{{{\left( {\sum {{x_i}} } \right)}^2}}}{n}}}{{n - 1}}\\ &=& \frac{{394913.6 - \frac{{{{\left( {2176.9} \right)}^2}}}{{12}}}}{{12 - 1}}\\ &=& 0.5245\end{array}\)

Thus, the sample variance for the provided data is 0.5245掳C square.