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Data Session 5: Solutions
 
Session 5 Part A Part B Part C Part D Part E Homework
 
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A B C D E
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Solutions for Session 5, Part E

See solutions for Problems: E1 | E2 | E3 | E4 | E5 | E6| E7| E8| E9| E10


Problem E1

By inspection, it is clear that Line Plot B has the least variation and Line Plot C has the most variation. Remember that the variation is from the mean, not between the values themselves, which is why Line Plot C has the most variation.

<< back to Problem E1


 

Problem E2

Number of Coins in Stack (x)

Deviation from the Mean (x-5)

Absolute Deviation from the Mean |x-5|

4

-1

1

4

-1

1

5

0

0

5

0

0

5

0

0

5

0

0

5

0

0

6

+1

1

6

+1

1

______

______

______

45

0

4

The MAD is 4 / 9, or approximately 0.44. This is much smaller than the MAD for Line Plot A, which indicates that the values in this allocation are much more closely grouped around the mean.

<< back to Problem E2


 

Problem E3

Number of Coins in Stack (x)

Deviation from the Mean (x-5)

Absolute Deviation from the Mean |x-5|

1

-4

4

1

-4

4

1

-4

4

1

-4

4

1

-4

4

10

+5

5

10

+5

5

10

+5

5

10

+5

5

______

______

______

45

0

40

The MAD is 40 / 9, or approximately 4.44. This is more than twice the MAD for Line Plot A and 10 times as large as the MAD for Line Plot B. The much larger MAD indicates that the values of Line Plot C are very far from the mean, as compared to the other two.

<< back to Problem E3


 

Problem E4

Answers will vary. Here is one possible line plot:

<< back to Problem E4


 

Problem E5

Answers will vary. Here is one possible line plot:

<< back to Problem E5


 

Problem E6

Answers will vary. Here is one possible line plot:

<< back to Problem E6


 

Problem E7

The reason it is impossible is that the MAD is the total of all absolute deviations. You may have noticed that in these problems the MAD is the sum of the deviations divided by 9. For the MAD to equal 1, the sum of the deviations would have to be exactly
9 (9 / 9 = 1). But the only way that could happen is if the total excess and the total deficit each were equal to 4.5. This would require splitting the coins, which cannot be done.

<< back to Problem E7


 

Problem E8

Number of Coins in Stack (x)

Deviation from the Mean
(x - 5)

Squared Deviation from the Mean
(x - 5)2

4

-1

1

4

-1

1

5

0

0

5

0

0

5

0

0

5

0

0

5

0

0

6

+1

1

6

+1

1

______

______

______

45

0

4

The variance is 4 / 9, or approximately 0.44.

The standard deviation is the square root of the variance, which is 2/3, or approximately 0.67. The standard deviation is slightly higher than the MAD (which is 0.44), and is significantly smaller than the standard deviation of Line Plot A (which is 2.05). The great difference in the standard deviations indicates that the values in Line Plot B are much more closely distributed around the mean.

<< back to Problem E8


 

Problem E9

Number of Coins in Stack (x)

Deviation from the Mean
(x - 5)

Squared Deviation from the Mean
(x - 5)2

1

-4

16

1

-4

16

1

-4

16

1

-4

16

1

-4

16

10

5

25

10

5

25

10

5

25

10

5

25

______

______

______

45

0

180

The variance for Line Plot B is 180 / 9 = 20.

The standard deviation -- the square root of 20 -- is approximately 4.47. This is very close to the MAD calculated in Problem E3 (4.44), and is much higher than the standard deviations for Line Plot A (2.05) and Line Plot B (0.67).

<< back to Problem E9


 

Problem E10

a. 

The mean would increase by 3.

b. 

The MAD would not change. Since the values in the list are each 3 larger, and the mean is also 3 larger, the deviations from the mean would remain the same.

c. 

The variance would not change, since it depends only on the deviation from the mean, not the values themselves. Since the mean increases by 3 along with the rest of the data set, none of the deviations will change.

d. 

Since the standard deviation is the square root of the (unchanged) variance, it will not change.

e. 

The mean would be doubled.

f. 

The MAD would be doubled, since all the deviations are now doubled, and the MAD is the average of these deviations.

g. 

The variance would be multiplied by 4. Since calculating the variance involves squaring the deviations, the newly doubled deviations would all be squared, resulting in values that are four times as large. For example, if a deviation was (+3), it now becomes (+6). The value used in the variance calculation changes from 32 = 9 to 62 = 36, which is four times as large.

h. 

The standard deviation would be doubled, since it is the square root of the variance.

<< back to Problem E10


 

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