Chapter 10.5, Problem 6CYU

i.

To determine

Explanation of Solution

Given information:

Tip(Miguel):14,68,52,21,63,32,43,35,70,37,32,16,47,38,48

Tip(Stephanie):34,52,43,39,41,50,46,36,37,47,39,49,44,36,50

Graph:

Miguel’s tip($):

  

Stephanie’s tip($):

  

ii.

To determine

To compare the data

Answer to Problem 6CYU

Miguel’s:

Maximum=70

Minimum=14

Median=38

Lower quartile=32

Upper quartile=52

Stephanie’s:

Maximum=52

Minimum=34

Median=41

Lower quartile=37

Upper quartile=49

Explanation of Solution

Given information:

Tip(Miguel):14,68,52,21,63,32,43,35,70,37,32,16,47,38,48

Tip(Stephanie):34,52,43,39,41,39,41,50,46,36,37,47,39,49,44,36,50

Formula used:

Lower quartile= $\frac{(N+1)}{4}$

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

Upper quartile= $\frac{3(N+1)}{4}$

Where N is number of tiles/data points

Calculation:

Miguel’s :

Sortingdata,

14,16,21,32,32,35,37,38,43,47,48,52,63,68,70

The 5 number summary is used because, the data can have abnormalities/distortion.

Maximum=70

Minimum=14

Substituting values,

Median is,

  ${(\frac{15+1}{2})}^{th}ob=8^{th}ob=38$

Lower quartile= $\frac{{(N+1)}^{th}ob}{4}=\frac{{(15+1)}^{th}ob}{4}=4^{th}ob=32$

Upper quartile= $\frac{3{(15+1)}^{th}ob}{4}=3\times 4^{th}ob={12}^{th}ob=52$

Stephanie’s:

Sorting data:

34,36,36,37,39,39,,41,43,44,46,47,49,50,50,52

Maximum=34

Minimum=52

Substituting values,

Median is,

  ${(\frac{15+1}{2})}^{th}ob=8^{th}ob=$ 41

Lower quartile= $\frac{{(N+1)}^{th}ob}{4}=\frac{{(15+1)}^{th}ob}{4}=4^{th}ob$ =37

Upper quartile= $\frac{3{(15+1)}^{th}ob}{4}=3\times 4^{th}ob={12}^{th}ob$ =49