91影视

The data for the alcohol content (%) is provided.

The size of the sample is 10.

a.

Let x represents the values of alcohol content in terms of percentage.

The measures of center are mean and median.

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{14.8 + 14.5 + 16.1 + ... + 15}}{{10}}\\ \approx 14.88\end{array}\)

Thus, the sample mean is 14.88%.

To compute the value of median, the data is arranged in ascending order,

For the even number of observations, the median value is computed as,

\(\begin{array}{c}\tilde x &=& {\rm{average}}\;{\rm{of}}\;{\left( {\frac{n}{2}} \right)^{th}}{\rm{and}}\;{\left( {\frac{n}{2} + 1} \right)^{th}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& {\rm{average}}\;{\rm{of}}\;{\left( {\frac{{10}}{2}} \right)^{th}}{\rm{and}}\;{\left( {\frac{{10}}{2} + 1} \right)^{th}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& {\rm{average}}\;{\rm{of}}\;{\left( 5 \right)^{th}}{\rm{and}}\;{\left( 6 \right)^{th}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& \frac{{14.6 + 14.8}}{2}\\ &=& 14.7\end{array}\)

Thus, the median value is 14.70%.

The mean alcohol content is 14.88%. It implies that there is an average of 14.88% alcohol content in the sampled zinfandels.

The median of the alcohol content is 14.70%. It implies that half of the sampled zinfandels have alcohol content greater than 14.70% while the other half have lower than this value.

It can be observed that the mean value is greater than the median. This implies that the distribution is skewed.

b.

The sample mean is 14.88%.

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 represented as,

Substituting the values,

The sample variance is computed as,

\(\begin{array}{c}{s^2} &=& \frac{{\sum {{{\left( {{x_i} - \bar x} \right)}^2}} }}{{n - 1}}\\ &=& \frac{{7.536}}{{10 - 1}}\\ &=& \frac{{7.536}}{9}\\ &=& 0.837\end{array}\)

Therefore, the sample variance is 0.837 % square.

c.

Subtracting 13 from each observation, the data is as follows,

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 are as follows,

Substituting the values, we have,

\(\begin{array}{c}{s^2} &=& \frac{{{S_{xx}}}}{{n - 1}}\\ &=& \frac{{42.88 - \frac{{{{\left( {18.8} \right)}^2}}}{{10}}}}{{10 - 1}}\\ &=& 0.837\end{array}\)

Therefore, the sample variance is 0.837% square.