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The data on the blood mercury concentration (mg/g) for adult females near contaminated rivers in Virginia is provided.

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{0.20 + 0.22 + 0.25 + ... + 3.25}}{{15}}\\ &=& \frac{{18.55}}{{15}}\\ &=& 1.237\end{array}\)

Thus, the sample mean for x is 1.237 mg/g.

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

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

Thus, the median value of x is 0.56.

The sample mean represents the average of values

The sample median separates the data into equal halves.

The two values are different in magnitude because the distribution of values is not symmetric but positively skewed.

b.

10% trimmed mean is the mean value obtained after removing 10% of the extreme (largest and smallest value each) from the data.

In this case, 10% of 15 is 1.5. Thus, the following procedure is carried out to compute trimmed mean.

A 10% trimmed mean is computed by removing the smallest and the largest observation from the data.

Remove one extreme value from each end to obtain the data as follows,

The sample mean is computed as,

\(\begin{array}{c}{{\bar x}_{\left( {tr1} \right)}} &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{0.22 + 0.25 + ... + 3.07}}{{13}}\\ &=& 1.162\end{array}\)

Now, again remove the smallest and largest value from the remaining dataset.

The data after removing is as follows,

The sample mean is computed as,

\(\begin{array}{c}{{\bar x}_{\left( {tr2} \right)}} &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{0.25 + 0.3 + ... + 2.25}}{{11}}\\ &=& 1.0736\end{array}\)

Now, the 10% trimmed mean for the data is computed as,

\(\begin{array}{c}{{\bar x}_{tr}} &=& \frac{{{{\bar x}_{\left( {tr1} \right)}} + {{\bar x}_{\left( {tr2} \right)}}}}{2}\\ &=& \frac{{1.162 + 1.0736}}{2}\\ &=& 1.118\end{array}\)

Thus, the 10% trimmed mean is 1.118.

Therefore, it can be observed that the trimmed mean (1.118) falls between the median value (0.56) and the mean value (1.237).

The median value of x is 0.56.

The value that the observation .20 be increased without impacting the value of the sample median is computed as,

\(\begin{array}{c}\frac{{0.20}}{{0.56}} = 0.357\\ \approx 0.36\end{array}\)

Thus, the observation .20 can be increased to 0.36 without impacting the value of the sample median.