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

See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6| C7| C8


Problem C1

The arrangement of the coins should look like this:

Here is the corresponding line plot:

Note that there are nine dots, one for each stack, which are placed in the line plot according to the number of coins in each stack.

<< back to Problem C1


 

Problem C2

The arrangement of the coins should look like this:

Here is the corresponding line plot:

<< back to Problem C2


 

Problem C3

The arrangement of the coins should look like this:

Here is the corresponding line plot:

<< back to Problem C3


 

Problem C4

There are multiple solutions to each of these problems, so answers will vary. Here are some possible solutions:

a. Here is a different line plot with a mean equal to 5:

b. Here is a line plot with a mean equal to 5 that has exactly 2 stacks of 5 coins:

c. Here is a line plot with a mean equal to 5 but a median not equal to 5:

d. Here is a line plot with a mean equal to 5 that has no 5-coin stacks:

e. Here is a line plot with a mean equal to 5 that has two 5-coin stacks, 4 stacks with more than 5 coins, and 3 stacks with fewer than 5 coins:

f. Here is a line plot with a mean equal to 5 that has two 5-coin stacks, 5 stacks with more than 5 coins, and 2 stacks with fewer than 5 coins:

g. Here is a line plot with a mean equal to 5 that has two 5-coin stacks, two 10-coin stacks, and 5 stacks with fewer than 5 coins:

<< back to Problem C4


 

Problem C5

You would need to remove 3 coins from one of the stacks of 5, resulting in a stack of 2 coins. This would reset the mean, since adding 3 coins to a stack changed the mean.

<< back to Problem C5


 

Problem C6

Yes, you could remove 1 coin from one stack and 2 coins from another.

Alternately, you could remove 1 coin from each of three stacks. You could even add 1 coin to a stack and remove 4 from another. The only requirement is that you subtract a total of 3 coins from the total number of coins (to counterbalance the 3 coins you added previously).

<< back to Problem C6


 

Problem C7

Answers will vary. Here are some possible solutions:

 

You could add 4 coins to the stack of 3 to create a stack of 7 coins.

 

You could add 2 coins to each of 2 stacks of 5 to create 2 stacks of 7 coins each.

 

You could add 1 coin apiece to 4 of the stacks of 5 to create 4 stacks of 6 coins each.

The only requirement is that you must add 4 coins to the total number of coins to return the sum to 45, which will return the mean to 45 / 9 = 5.

<< back to Problem C7


 

Problem C8

As with Problem C4, answers will vary. See Problem C4 for some sample solutions.

<< back to Problem C8


 

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