The May 1, 2009 issue of ?The Montclarian ?reported the following home sale amounts for a sample of homes in Alameda, CA that were sold the previous month (1000s of $): a. Calculate and interpret the sample mean and median. b. Suppose the 6th observation had been 985 rather than 1285. How would the mean and median change? c. Calculate a 20% trimmed mean by first trimming the two smallest and two largest observations. d. Calculate a 15% trimmed mean.

Answer Step 1 of 4 a) Mean=X /n=(590+815+575+608+350+1285+408+540+555+679)/10=6405/10=640.5 Arrange the given values in the ascending order 350, 408, 540, 555, 575, 590,608, 679, 815, 1285 n+1 th Median= ( 2 ) value =( 10+1 ) value 2 =5.5 value =(575+590)/2 =582.5 Step 2 of 4 b)Suppose the 6th observation had been 985 rather than 1285 Mean=X /n=(590+815+575+608+350+985+408+540+555+679)/10=6105/10=610.5 Arrange the given values in the ascending order 350, 408, 540, 555, 575, 590,608, 679, 815, 985 n+1 th Median= ( 2 ) value 10+1 th =( 2 ) value =5.5 value =(575+590)/2 =582.5