Answer:
Median: 14.6, Q1: 6.1, Q3: 27.1, IR: 21, outliers: none
Step-by-step explanation:
Step 1: order the data from the least to the largest.
2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, 14.6, 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5
Step 2: find the median.
The median is the middle value, which is the 8th value in the data set.
2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, [14.6,] 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5
Median = 14.6
Step 2: Find Q1,
Q1 is the middle value of the lower part of the data set that is divided by the median to your left.
2.8, 3.9, 5.3, (6.1), 6.5, 7.1, 12.5, [14.6], 16.4, 16.4, 20.8, 27.1, 28.1, 30.9, 53.5
Q1 = 6.1
Step 3: find Q3.
Q3 is the middle value of the upper part of the given data set.
2.8, 3.9, 5.3, 6.1, 6.5, 7.1, 12.5, [14.6], 16.4, 16.4, 20.8, (27.1), 28.1, 30.9, 53.5
Q3 = 27.1
Step 4: find interquartile range (IR)
IR = Q3 - Q1 = [tex] 27.1 - 6.1 = 21 [/tex]
Step 5: check if there is any outlier.
Formula for checking for outlier = [tex] Q1 - 1.5*IR [/tex]
Then compare the result you get with the given values in the data set. Any value in the data set that is less than the result we get is considered an outlier.
Thus,
[tex] Q1 - 1.5*IR [/tex]
[tex]6.1 - 1.5*21 = -25.4[/tex]
There are no value in the given data set that is less than -25.4. Therefore, there is no outlier.
