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Given the following data:

405, 410, 424, 426, 505, 505, 575, 575, 600, 625, 865
a. Find the four quartiles
Q1-
Q2™
Q3=
(aka median)
Q4=
b. Find the IQR interquartile range (Q3 - Q1)
(max value)
c. Find the lower boundary (Q1 - 1.5(IQR))
e. Find the outlier
d. Find the upper boundary (Q3 +1.5(IQR))

User MKa
by
4.3k points

1 Answer

4 votes

Answer:

To find the quartiles, we first need to arrange the data in ascending order:

405, 410, 424, 426, 505, 505, 575, 575, 600, 625, 865

To find the first quartile (Q1), we take the average of the middle two numbers of the lower half of the data:

(405 + 410) / 2 = 407.5

To find the second quartile (Q2), we take the average of the two middle numbers of the data:

(424 + 426) / 2 = 425

To find the third quartile (Q3), we take the average of the middle two numbers of the upper half of the data:

(575 + 575) / 2 = 575

To find the fourth quartile (Q4), we take the average of the two highest numbers of the data:

(600 + 625) / 2 = 612.5

Therefore, the quartiles are Q1 = 407.5, Q2 = 425, Q3 = 575, and Q4 = 612.5.

To find the interquartile range (IQR), we subtract Q1 from Q3:

IQR = Q3 - Q1 = 575 - 407.5 = 167.5

To find the lower boundary, we subtract 1.5 times the IQR from Q1:

lower boundary = Q1 - 1.5 * IQR = 407.5 - 1.5 * 167.5 = 196.25

To find the upper boundary, we add 1.5 times the IQR to Q3:

upper boundary = Q3 + 1.5 * IQR = 575 + 1.5 * 167.5 = 783.75

Any number that is less than the lower boundary or greater than the upper boundary is considered an outlier. In this case, the only outlier is 865, which is greater than the upper boundary.

User Steve Bosman
by
4.4k points