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The data on the concentration (EU/mg) in settled dust for one sample of urban homes and another of farm homes are provided.

Let x represents the concentration (EU/mg) in settled dust for one sample of urban homes.

Let y represent the concentration (EU/mg) in settled dust for another farm homes.

The sample mean is computed as,

For x,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{6 + 5 + 11 + ... + 23}}{{11}}\\ &=& \frac{{237}}{{11}}\\ &=& 21.5\end{array}\)

Thus, the sample mean for x is 21.5EU/mg.

For y,

\(\begin{array}{c}\bar y &=& \frac{{\sum {{y_i}} }}{{{n_y}}}\\ &=& \frac{{4 + 14 + 11 + ... + 0.3}}{{15}}\\ &=& \frac{{128.4}}{{15}}\\ &=& 8.6\end{array}\)

Thus, the sample mean for y is 8.6EU/mg.

Therefore, it can be concluded that the average concentration (EU/mg) in settled dust for one sample of urban homes is more than compared to the another of farm homes.

Let x represents the concentration (EU/mg) in settled dust for one sample of urban homes.

Let y represent the concentration (EU/mg) in settled dust for another farm homes.

The sample median is computed by first ordering the data in ascending order.

For x,

The data is arranged as,

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

\(\begin{array}{c}\tilde x &=& {\left( {\frac{{n + 1}}{2}} \right)^{th}}ordered\;value\\ &=& {\left( {\frac{{11 + 1}}{2}} \right)^{th}}ordered\;value\\ &=& {\left( 6 \right)^{th}}ordered\;value\end{array}\)

Thus, the median value of x is 17EU/mg.

For y,

The data is arranged as,

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

\(\begin{array}{c}\tilde y &=& {\left( {\frac{{n + 1}}{2}} \right)^{th}}ordered\;value\\ &=& {\left( {\frac{{15 + 1}}{2}} \right)^{th}}ordered\;value\\ &=& {\left( 8 \right)^{th}}ordered\;value\end{array}\)

Thus, the median value of y is 8.9EU/mg.

Therefore, it can be concluded that the median concentration (EU/mg) in settled dust for one sample of urban homes is more than compared to the another of farm homes.

The median for the urban sample is so different from the mean for that sample because mean represents the average and the median value represents the centre.

Let x represents the concentration (EU/mg) in settled dust for one sample of urban homes.

Let y represent the concentration (EU/mg) in settled dust for another farm homes.

The trimmed mean for each sample is obtained by deleting few of the smallest and largest observation.

The trimmed data for x is,

The sample mean is computed as,

For x,

\(\begin{array}{c}{{\bar x}_{tr}} &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{5 + 5 + 6 + ... + 35}}{9}\\ &=& \frac{{153}}{9}\\ &=& 17\end{array}\)

Thus, the trimmed mean for x is 17EU/mg.

The trimmed data for y is,

For y,

\(\begin{array}{c}{{\bar y}_{\left( {tr} \right)}} &=& \frac{{\sum {{y_i}} }}{{{n_y}}}\\ &=& \frac{{2 + 3 + 4 + ... + 20}}{{13}}\\ &=& \frac{{107.1}}{{13}}\\ &=& 8.2\end{array}\)

Thus, the trimmed mean for y is 8.2EU/mg.

The trimmed percentages (P) are computed using the following formula for each of the two samples.

\(\frac{P}{{100}}*N = K \Rightarrow {\bf{P = }}\frac{{\bf{K}}}{{\bf{N}}}*{\bf{100}}\)

Where N is the total number of observations and K is the number of values removed from each end.

In case of urban homes,

\(\begin{array}{l}N = 11\\K = 1\end{array}\)

The trimmed percentage is,

\(\begin{array}{c}{P_U} &=& \frac{1}{{11}}*100\\ &=& 9.09\% \end{array}\)

In case of farm homes,

\(\begin{array}{l}N = 15\\K = 1\end{array}\)

The trimmed percentage is,

\(\begin{array}{c}{P_F} &=& \frac{1}{{15}}*100\\ &=& 6.67\% \end{array}\)

On comparison of trimmed mean with the mean of the complete data, it is concluded that the trimmed mean is smaller in magnitude for both urban and farm homes.

Similarly, it is also observed that the trimmed mean is lower in magnitude than the median values.