select the outlier in the data set.\n602 558 560 567 575 607 563 510 629 617 5735 505 606 570\n509 574 599 517

Explanation:

Step1: Arrange data in ascending order

505, 509, 510, 517, 558, 560, 563, 567, 570, 573, 574, 575, 599, 602, 606, 607, 617, 629

Step2: Calculate quartiles

First, find the median (Q2). Since there are 18 data - points, the median is the average of the 9th and 10th ordered values. Q2=$\frac{570 + 573}{2}=571.5$. The lower half of the data is 505, 509, 510, 517, 558, 560, 563, 567, 570. The median of the lower half (Q1) is the 5th value, so Q1 = 558. The upper half of the data is 573, 574, 575, 599, 602, 606, 607, 617, 629. The median of the upper half (Q3) is the 5th value of the upper - half, so Q3 = 602.

Step3: Calculate the inter - quartile range (IQR)

IQR = Q3 - Q1 = 602 - 558 = 44.

Step4: Determine the outlier boundaries

Lower boundary = Q1-1.5IQR = 558 - 1.544 = 558 - 66 = 492. Upper boundary = Q3 + 1.5IQR = 602+1.544 = 602 + 66 = 668.

Step5: Identify the outlier

All the data points 505, 509, 510, 517, 558, 560, 563, 567, 570, 573, 574, 575, 599, 602, 606, 607, 617, 629 are within the range [492, 668]. There is no outlier in this data - set.

Answer:

There is no outlier.