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The data are provided consists of 50 observations, so the sample size is 50.

Let \(\tilde x\) be the required sample median.

Since \({\bf{n = 50}}\) is even, the sample median is the average of \({\left( {\frac{{\bf{n}}}{{\bf{2}}}} \right)^{{\bf{th}}}}\) and \({\left( {\frac{{\bf{n}}}{{\bf{2}}}{\bf{ + 1}}} \right)^{{\bf{th}}}}\)term.

Calculate the sample median as follows,

\(\begin{array}{c}\tilde x &=& {\rm{average}}\,{\rm{of}}\,{\left( {\frac{n}{2}} \right)^{th}}{\rm{and}}{\left( {\frac{n}{2} + 1} \right)^{th}}{\rm{ordered}}\,{\rm{values}}\\ &=& \frac{{{x_{25}} + {x_{26}}}}{2}\end{array}\)

Substitute the values of\({x_{25}} = 91\)and\({x_{26}} = 93\)in the above formula,

\(\begin{array}{c}\tilde x &=& \frac{{91 + 93}}{2}\\ &=& \frac{{184}}{2}\\ &=& 92\end{array}\)

Thus, the sample median is 92.

Let \({\bar x_{Tr}}\)be the required trimmed mean.

The number of extreme values to be trimmed are,

\(\begin{array}{c}50\left( {0.25} \right) = 12.5\\ \approx 13\end{array}\)

Thus, the smallest and largest 13 values of the data set are removed from the data set.

The remaining data for 24 valuesis as follows,

67, 68, 71, 74, 76, 78, 79, 81, 84, 85, 89, 91, 93, 96, 99, 101, 104, 105, 105, 112, 118, 123, 136, 139

The 25% trimmed mean is computed as follows,

\({\bar x_{Tr\left( {25} \right)}} = \frac{{\sum {{x_{Tr}}} }}{{{n_{Tr}}}}\)

Substitute the value of\({n_{Tr}} = 24\)in the above formula,

\(\begin{array}{c}{{\bar x}_{Tr\left( {25} \right)}} &=& \frac{{\left( {67 + 68 + 71 + 74 + \ldots + 123 + 136 + 139} \right)}}{{24}}\\ &=& \frac{{2274}}{{24}}\\ &=& 94.75\end{array}\)

Thus, the 25% trimmed mean is 94.75.

Let\({\bar x_{Tr}}\)be the required trimmed mean.

The number of extreme values to be trimmed are,

\(50\left( {0.10} \right) = 5\)

Thus, the smallest and largest 5 values of the data set are removed from the data set.

The remaining data for 24 values is as follows,

36, 39, 44, 47, 50, 59, 61, 65, 67, 68, 71, 74, 76, 78, 79, 81, 84, 85, 89, 91, 93, 96, 99, 101, 104, 105, 105, 112, 118, 123, 136, 139, 141, 148, 158, 161, 168, 184, 206, 248

The trimmed mean is,

\({\bar x_{Tr\left( {10} \right)}} = \frac{{\sum {{x_{Tr}}} }}{{{n_{Tr}}}}\)

Substitute the value of\({n_{Tr}} = 40\)in the above formula,

\(\begin{array}{c}{{\bar x}_{Tr\left( {10} \right)}} &=& \frac{{\left( {36 + 39 + 44 + \ldots + 184 + 206 + 248} \right)}}{{40}}\\ &=& \frac{{4089}}{{40}}\\ &=& 102.225\end{array}\)

Thus, the 10% trimmed mean is 102.225.

Let\(\bar x\)be the required sample mean.

There are 50 observations, that is,\(n = 50\)

Sample mean formula:

\(\bar x = \frac{{\sum {{x_i}} }}{n}\)

Substitute the value of n and\(\sum {{x_i}} \)in the above formula,

\(\begin{array}{c}\bar x &=& \frac{{\left( {11 + 14 + 20 + \ldots + 322 + 388 + 513} \right)}}{{50}}\\ &=& \frac{{5963}}{{50}}\\ &=& 119.26\end{array}\)

Thus, the sample mean is 119.26.

As the value of sample mean is 119.26 higher than the obtained value of median, the distribution of 50 observations becomes skewed to the right or positively skewed.

The highest value is observed for sample mean as compared to all other measures. Also, as the percentage of trimmed mean increases, the value of mean decreases.