Question

The May 1, 2009, issue of the Mont clarian reported the following home sale amounts for a sample of homes in Alameda, CA that were sold the previous month (1000s of $):

590 815 575 608 350 1285 408 540 555 679

1. Calculate and interpret the sample mean and median.
2. Suppose the 6th observation had been 985 rather than 1285. How would the mean and median change?
3. Calculate a 20% trimmed mean by first trimming the two smallest and two largest observations.
4. Calculate a 15% trimmed mean.

Short answer

a. The sample mean amount is 640.5(1000s of $) and median value is 582.5(1000s of $).

b. The sample mean amount is 610.5(1000s of $) and The median value is 582.5(1000s of $).

c. The 20% trimmed mean is 591.2. (1000s of $)

d. The 15% trimmed mean is 593.71(1000s of $).

Step-by-step solution

Given information

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

Compute the sample mean and median

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.

350
408
540
555
575
590
608
679
815
1285

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 $).

Interpret the values

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 $).

Compute the sample mean and median by replacing a value

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,

350

408

540

555

575

590

608

679

815

985

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.

Compute the 20% trimmed mean

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,

540

555

575

590

608

679

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 $).

Compute the 15% trimmed mean

d. Let x represents the home sales amounts.

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

408

540

555

575

590

608

679

815

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,

540

555

575

590

608

679

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 $).