Читать книгу Statistics and Probability with Applications for Engineers and Scientists Using MINITAB, R and JMP - Bhisham C. Gupta, Irwin Guttman - Страница 71

1 Step 1. Write the data in an ascending order and rank them from 1 to 16, since .Observations in ascending order78101212151516171819202225250300Ranks12345678910111213141516

2 Step 2. Rank of the median .

3 Step 3. Following our previous discussion, the median in this case is the average of the values corresponding to their ranks of 8 and 9. Thus, the median of this data is . In other words, the median sales of the given individuals is $16,500. We remark that eight observations fall below 16.5, and eight observations fall above 16.5.

It is important to note that the median may or may not be one of the values of the data set as in this case. Whenever the sample size is odd, the median is the center value, and whenever it is even, the median is always the average of the two middle values when the data are arranged in the ascending order.

Finally, note that the data in this example contain the two values $250,000 and $300,000. These large values seem to be the sales of top‐performing sales personnel and may be considered as outliers. In this case, the mean of these data is

Note that the mean of 47.875 is much larger than the median of 16.5. It is obvious that the mean of these data has been adversely affected by the outliers. Hence, in this case, the mean does not adequately represent the measure of centrality of the data set, so that the median would more accurately identify the location of the center of the data.

Furthermore, if we replace the extreme values of 250 and 300, for example, by 25 and 30, respectively, then the median will not change, whereas the mean becomes 16.937, namely $16,937. Thus, the new data obtained by replacing the values 250 and 300 with 25 and 30, respectively, do not contain any outliers. The new mean value is more consistent with the true average sales.