Instead of considering a data value to be an outlier if it is \"very far away fr
ID: 3242498 • Letter: I
Question
Instead of considering a data value to be an outlier if it is "very far away from almost all of the other data values" consider an outlier to be a value that is above Q, by an amount an amount greater than 1.5 times IQR. Use the data set giver below to find (a) the 5.number summary, (b) the interquartile range (IQR), and (c) any outliers. 2 3 4 5611 14 1515 17 22 32 50 a. The 5-number summary is. (Type exact answers. Use ascending order.) b. The interquartile range is. c The outliers are. (Use a comma to separate answers as needed.)Explanation / Answer
Solution:-
a) The five number summary is 2, 4.5, 14, 19.5, 50.
Minimum = 2
First quartile = 4.5
Median = 14
Third quartile = 19.5
Maximum = 50
b) IQR = 15
IQR = Third quartile - First quartile
IQR = 19.5 - 4.5
IQR = 15
c) The outliers are 32 and 50.
Upper and lower bond for outlier are above Q1 by 1.5 × IQR and below Q1 by 1.5 × IQR.
IQR × 1.5 = 1.5 × 15 = 22.5
Upper Bond = 4.5 + 22.5 = 27.0
Outliers are 32 and 50.