Chapter 10.5, Problem 3BGP

To determine

To compare the data sets given.

Answer to Problem 3BGP

For set 1

Median=13

Lower quartile=9

Upper quartile=22

Minimum=4

Maximum=33

For set 2

Median=19

Lower quartile=13

Upper quartile=21

Minimum=4

Maximum=26

Explanation of Solution

Given information:

Two data sets are:

set 1 :8,4,18,7,13,26,12,6,20,5,9,24,8,16,31,13,17,10,8,22,12,25,13,11,18,12,16,22,25,33

set 2 :21,4,20,13,22,6,10,23,13,25,14,16,19,21,19,8,20,18,9,14,21,17,19,22,4,19,21,26

5 number summary is used because the data can have distortions/abnormalities and 5 number summary is least affected by outliers

Calculation:

For the 1^st data set,

4,5,6,7,8,8,8,9,10,11,12,12,12,13,13,13,16,16,17,18,18,20,22,22,24,25,25,26,31,33

Total data points: 30

Minimum is 4.

Maximum is 33.

Median= $\frac{{(\frac{N}{2})}^{th}ob+{(\frac{N}{2}+1)}^{th}ob}{2}=\frac{{(\frac{30}{2})}^{th}+{(\frac{30}{2}+1)}^{th}}{2}=\frac{{15}^{th}+{16}^{th}}{2}=\frac{13+13}{2}=13$

Pair 15^th and 16^th terms.

Lower half median gives lower quartile, include the paired term.

Total terms in lower term=15

Lower Quartile= $\frac{(N+1)}{2}=\frac{(15+1)}{2}=8^{th}ob=9$

Upper half median gives upper quartile, include the paired term. Total terms in upper half=15

Upper quartile= $\frac{(N+1)}{2}=\frac{(15+1)}{2}=8^{th}ob=22$

For set 2:

After sorting,

4,4,6,8,9,10,13,13,14,14,16,17,18,19,19,19,19,20,20,21,21,21,21,22,22,23,25,26

Total terms=28

Median= $\frac{{(\frac{N}{2})}^{th}ob+{(\frac{N}{2}+1)}^{th}ob}{2}=\frac{{(\frac{28}{2})}^{th}+{(\frac{28}{2}+1)}^{th}}{2}=\frac{{14}^{th}+{15}^{th}}{2}=\frac{19+19}{2}=19$

Pair 15^th and 14^th terms.

Lower half median gives lower quartile, include the paired term.

Total terms in lower term=15

Lower quartile= $\frac{(N+1)}{2}=\frac{(15+1)}{2}=8^{th}ob=13$

Upper half median gives upper quartile, include the paired term.

Total terms in lower term=15

Upper quartile= $\frac{(N+1)}{2}=\frac{(15+1)}{2}=8^{th}ob=21$

Minimum=4

Maximum=26