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The data for home sale amounts in (1000s of $) for a sample of homes in Alameda, CA, sold the previous month is provided.

a. Let x represents the home sales amounts.

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{590 + 815 + 575 + ... + 679}}{{10}}\\ &=& \frac{{6405}}{{10}}\\ &=& 640.5\end{array}\)

Thus, the sample mean amount is 640.5(1000s of $).

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

The data provided is arranged in ascending order.

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

\(\begin{array}{c}\tilde x &=& {\rm{average}}\;{\rm{of}}\;{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{and}}\;{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& {\rm{average}}\;{\rm{of}}\;{\left( {\frac{{{\rm{10}}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{and}}\;{\left( {\frac{{{\rm{10}}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& {\rm{average}}\;{\rm{of}}\;{\left( {\rm{5}} \right)^{{\rm{th}}}}{\rm{and}}\;{\left( {\rm{6}} \right)^{{\rm{th}}}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& \frac{{575 + 590}}{2}\\ &=& 582.5\end{array}\)

Thus, the median value is 582.5(1000s of $).

Interpretation of mean:

The averagehome sale amount for a sample of homes in Alameda, CA that were sold the previous month is 640.5 (1000s of $).

Interpretation of median:

The central or middlemosthome sale amount for a sample of homes in Alameda, CA that were sold the previous month is582.5 (1000s of $).

b. The value of the 6^th observation is 985 instead of 1285.

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{590 + 815 + 575 + ... + 679}}{{10}}\\ &=& \frac{{6105}}{{10}}\\ &=& 610.5\end{array}\)

Thus, the sample mean amount is 610.5(1000s of $).

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

The data is arranged as,

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

\(\begin{array}{c}\tilde x &=& {\rm{average}}\;{\rm{of}}\;{\left( {\frac{{\rm{n}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{and}}\;{\left( {\frac{{\rm{n}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& {\rm{average}}\;{\rm{of}}\;{\left( {\frac{{{\rm{10}}}}{{\rm{2}}}} \right)^{{\rm{th}}}}{\rm{and}}\;{\left( {\frac{{{\rm{10}}}}{{\rm{2}}}{\rm{ + 1}}} \right)^{{\rm{th}}}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& {\rm{average}}\;{\rm{of}}\;{\left( {\rm{5}} \right)^{{\rm{th}}}}{\rm{and}}\;{\left( {\rm{6}} \right)^{{\rm{th}}}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& \frac{{575 + 590}}{2}\\ &=& 582.5\end{array}\)

Thus, the median value is 582.5(1000s of $).

Therefore, with the change in the value of 6^th observation the mean value decreases from 640.5 to 610.5 while the median value remains the same.

c. Let x represents the home sales amounts.

Trimmed mean is obtained by omitting few smallest and largest values from the data.

The data after trimming the two smallest and two largest observations is as follows,

The 20% trimmed mean is computed as,

\(\begin{array}{c}{{\bar x}_{\left( {tr20} \right)}} &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{540 + 555 + 575 + ... + 679}}{6}\\ &=& \frac{{3547}}{6}\\ &=& 591.2\end{array}\)

Thus, the 20% trimmed mean is 591.2(1000s of $).

A 15% trimmed mean is computed in two steps as shown below; by removing the smallest and the largest observation from the data as 15% of 10 is 1.5.

The data after removing the smallest and largest observation is as follows

The sample mean for the remaining values is computed as,

\(\begin{array}{c}{{\bar x}_{\left( {tr15} \right)}} &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{408 + 540 + 555 + ... + 815}}{8}\\ &=& 596.25\end{array}\)

Thus, the sample meanof the remaining valuesis 591.2 (1000s of $).

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

The data after removing is as follows,

The sample meanfor the remaining values is computed as,

\(\begin{array}{c}{{\bar x}_{\left( {tr2} \right)}} &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{540 + 555 + ... + 679}}{6}\\ &=& 591.167\end{array}\)

Thus, the sample meanof the remaining valuesis 591.167(1000s of $).

Now, the trimmed mean is computed as,

\(\begin{array}{c}{{\bar x}_{tr}} &=& \frac{{{{\bar x}_{\left( {tr20} \right)}} + {{\bar x}_{\left( {tr15} \right)}}}}{2}\\ &=& \frac{{596.25 + 591.167}}{2}\\ &=& 593.708\\ \approx 593.71\end{array}\)

Thus, the 15% trimmed mean is 593.71(1000s of $).