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Data Session 4: Solutions
 
Session 4 Part A Part B Part C Part D Part E Homework
 
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Solutions for Session 4, Part E

See solutions for Problems: E1 | E2 | E3 | E4 | E5 | E6| E7


Problem E1

Here are some examples. If there were 15 raisins (n = 15), the median would be in position (8). If n = 16, the median would be in position (8.5). If n = 17, the median would be in position (9). A general mathematical rule is that the position of the median is (n + 1) / 2, where n is the number of items in the list.

<< back to Problem E1


 

Problem E2

Since six is an even number, this is a case where you would need to draw a line to represent the position of Q1, the median of the six noodles to the left of Med. Using the formula (n + 1) / 2 from Problem E1 gives us (6 + 1) / 2 = 7 / 2 = 3.5. Therefore, position (3.5L), halfway between positions (3L) and (4L) from the low end of the noodles, is the position (though not the value) of Q1.

<< back to Problem E2


 

Problem E3

Again, you'll need to draw a line to represent the position of Q3. As in Problem E2, the formula (n + 1) / 2 gives us (6 + 1) / 2 = 7 / 2 = 3.5. Therefore, position (3.5R), halfway between positions (3R) and (4R) from the noodles, is the position (though not the value) of Q3.

<< back to Problem E3


 

Problem E4

Here is the Five-Number Summary:

 

Min is in position (1L); Min = 13.

 

Max is in position (1R); Max = 127.

 

Med is in position (13 + 1) / 2 = (7); Med = 74.

 

There are six positions to the left of (7), so Q1 is in position (6 + 1) / 2 = (3.5L). The value of Q1 is (28 + 33) / 2; Q1 = 30.5.

 

There are six positions to the right of (7), so Q3 is in position (6 + 1) / 2 = (3.5R). The value of Q3 is (102 + 118) / 2; Q3 = 110.

<< back to Problem E4


 

Problem E5

First, number the positions as you did in Problem E4. The center position will be marked with an (8). Here is the Five-Number Summary:

 

The minimum is in position (1L); Min = 10.

 

The maximum is in position (1R); Max = 89.

 

The median is in position (15 + 1) / 2 = (8); Med = 26.

 

The first quartile is in position (7 + 1) / 2 = (4L); Q1 = 18.

 

The third quartile is in position (4R); Q3 = 51.

<< back to Problem E5


 

Problem E6

Again, number the positions as you did in Problem E4. This time, there will be two values numbered (10) in the center of the ordered list. Here is the Five-Number Summary:

 

The minimum is in position (1L); Min = 1.

 

The maximum is in position (1R); Max = 53.

 

The median is in position (20 + 1) / 2 = (10.5), which means it is the average of the two values numbered (10), or (17 + 20) / 2; Med = 18.5.

 

The first quartile is in position (10 + 1) / 2 = (5.5L), which is (4 + 4) / 2; Q1 = 4.

 

The third quartile is in position (5.5R), which is (34 + 38) / 2; Q3 = 36.

<< back to Problem E6


 

Problem E7

The ordered list is as follows:

Length of Needles (in millimeters)

37

40

43

47

48

49

56

64

68

69

71

86

93

93

109

109

110

111

117

120

a. Here is the Five-Number Summary:
Min = 37
Q1 = 48.5
Med = 70
Q3 = 109
Max = 120

b. Here is the box plot:

c. Based on these measurements:

 

All pine needles have lengths between 37 mm and 120 mm.

 

Approximately half the pine needles have lengths less than 70 mm.

 

Approximately half the pine needles have lengths greater than 70 mm.

 

Approximately half the pine needles have lengths between 48.5 mm and 109 mm.

 

The widest range of needle lengths seems to be in the third quartile, where 25% of the needles are between 70 mm and 109 mm.

 

The longest and shortest needles fall in very tight ranges; the longest 25% of needles are between 109 and 120 mm, and the shortest 25% are between 37 mm and 48.5 mm.

<< back to Problem E7


 

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