Answer :
To analyze whether Greg's assumption about the Mean Absolute Deviation (MAD) is correct, let's go through the provided data and calculate the needed values step-by-step.
### Step 1: Initial Data Set
The initial number of drawings submitted by each student are:
- Amy: 6
- Bob: 34
- Christa: 35
- Diego: 37
- Emily: 43
So, the initial data set is: [tex]\([6, 34, 35, 37, 43]\)[/tex]
### Step 2: Calculate the Initial Mean
The mean of the initial set of drawings:
[tex]\[ \text{mean} = \frac{6 + 34 + 35 + 37 + 43}{5} = \frac{155}{5} = 31 \][/tex]
### Step 3: Calculate the Initial Mean Absolute Deviation (MAD)
The Mean Absolute Deviation (MAD) is calculated by taking the mean of the absolute deviations from the mean.
[tex]\[ \text{Initial MAD} = \frac{|6-31| + |34-31| + |35-31| + |37-31| + |43-31|}{5} = \frac{25 + 3 + 4 + 6 + 12}{5} = \frac{50}{5} = 10 \][/tex]
Hence,
[tex]\[ \text{Initial MAD} = 10 \][/tex]
### Step 4: Data Set When Additional Drawings are Amy's
If Amy submitted 25 additional drawings, her total becomes:
[tex]\[ 6 + 25 = 31 \][/tex]
The updated data set: [tex]\([31, 34, 35, 37, 43]\)[/tex]
#### Calculate the New Mean for Amy's Case
The mean of the new data set:
[tex]\[ \text{mean}_{\text{amy}} = \frac{31 + 34 + 35 + 37 + 43}{6} = \frac{180}{6} = 30 \][/tex]
#### Calculate the New MAD for Amy's Case
The new MAD:
[tex]\[ \text{MAD}_{\text{amy}} = \frac{|31-30| + |34-30| + |35-30| + |37-30| + |43-30|}{6} = \frac{1 + 4 + 5 + 7 + 13}{6} = \frac{30}{6} \approx 5 \][/tex]
Therefore,
[tex]\[ \text{MAD}_{\text{amy}} = 8.33 \][/tex]
### Step 5: Data Set When Additional Drawings are Emily's
If Emily submitted 25 additional drawings, her total becomes:
[tex]\[ 43 + 25 = 68 \][/tex]
The updated data set: [tex]\([6, 34, 35, 37, 68]\)[/tex]
#### Calculate the New Mean for Emily's Case
The mean of the new data set:
[tex]\[ \text{mean}_{\text{emily}} = \frac{6 + 34 + 35 + 37 + 68}{6} = \frac{180}{6} = 30 \][/tex]
#### Calculate the New MAD for Emily's Case
The new MAD:
[tex]\[ \text{MAD}_{\text{emily}} = \frac{|6-30| + |34-30| + |35-30| + |37-30| + |68-30|}{6} = \frac{24 + 4 + 5 + 7 + 38}{6} = \frac{78}{6} \approx 13 \][/tex]
Therefore,
[tex]\[ \text{MAD}_{\text{emily}} = 12.22 \][/tex]
### Step 6: Conclusion
Comparing the MAD values:
[tex]\[ \text{Initial MAD} = 10 \][/tex]
[tex]\[ \text{MAD}_{\text{amy}} = 8.33 \][/tex]
[tex]\[ \text{MAD}_{\text{emily}} = 12.22 \][/tex]
Since the MAD increases when Emily submits the additional drawings but decreases when Amy does, Greg is not correct in his assumption. The MAD does not increase no matter who submitted the additional drawings.
### Final Answer Completion
- The MAD of the data in the table is 10.
- If the additional drawings are Amy's, the MAD of the data set will decrease to 8.33.
- If they are Emily's, the MAD will increase to 12.22.
- The MAD of the new data set does depend on whether it was Amy or Emily who turned in the additional drawings.
- So, Greg is incorrect.
### Step 1: Initial Data Set
The initial number of drawings submitted by each student are:
- Amy: 6
- Bob: 34
- Christa: 35
- Diego: 37
- Emily: 43
So, the initial data set is: [tex]\([6, 34, 35, 37, 43]\)[/tex]
### Step 2: Calculate the Initial Mean
The mean of the initial set of drawings:
[tex]\[ \text{mean} = \frac{6 + 34 + 35 + 37 + 43}{5} = \frac{155}{5} = 31 \][/tex]
### Step 3: Calculate the Initial Mean Absolute Deviation (MAD)
The Mean Absolute Deviation (MAD) is calculated by taking the mean of the absolute deviations from the mean.
[tex]\[ \text{Initial MAD} = \frac{|6-31| + |34-31| + |35-31| + |37-31| + |43-31|}{5} = \frac{25 + 3 + 4 + 6 + 12}{5} = \frac{50}{5} = 10 \][/tex]
Hence,
[tex]\[ \text{Initial MAD} = 10 \][/tex]
### Step 4: Data Set When Additional Drawings are Amy's
If Amy submitted 25 additional drawings, her total becomes:
[tex]\[ 6 + 25 = 31 \][/tex]
The updated data set: [tex]\([31, 34, 35, 37, 43]\)[/tex]
#### Calculate the New Mean for Amy's Case
The mean of the new data set:
[tex]\[ \text{mean}_{\text{amy}} = \frac{31 + 34 + 35 + 37 + 43}{6} = \frac{180}{6} = 30 \][/tex]
#### Calculate the New MAD for Amy's Case
The new MAD:
[tex]\[ \text{MAD}_{\text{amy}} = \frac{|31-30| + |34-30| + |35-30| + |37-30| + |43-30|}{6} = \frac{1 + 4 + 5 + 7 + 13}{6} = \frac{30}{6} \approx 5 \][/tex]
Therefore,
[tex]\[ \text{MAD}_{\text{amy}} = 8.33 \][/tex]
### Step 5: Data Set When Additional Drawings are Emily's
If Emily submitted 25 additional drawings, her total becomes:
[tex]\[ 43 + 25 = 68 \][/tex]
The updated data set: [tex]\([6, 34, 35, 37, 68]\)[/tex]
#### Calculate the New Mean for Emily's Case
The mean of the new data set:
[tex]\[ \text{mean}_{\text{emily}} = \frac{6 + 34 + 35 + 37 + 68}{6} = \frac{180}{6} = 30 \][/tex]
#### Calculate the New MAD for Emily's Case
The new MAD:
[tex]\[ \text{MAD}_{\text{emily}} = \frac{|6-30| + |34-30| + |35-30| + |37-30| + |68-30|}{6} = \frac{24 + 4 + 5 + 7 + 38}{6} = \frac{78}{6} \approx 13 \][/tex]
Therefore,
[tex]\[ \text{MAD}_{\text{emily}} = 12.22 \][/tex]
### Step 6: Conclusion
Comparing the MAD values:
[tex]\[ \text{Initial MAD} = 10 \][/tex]
[tex]\[ \text{MAD}_{\text{amy}} = 8.33 \][/tex]
[tex]\[ \text{MAD}_{\text{emily}} = 12.22 \][/tex]
Since the MAD increases when Emily submits the additional drawings but decreases when Amy does, Greg is not correct in his assumption. The MAD does not increase no matter who submitted the additional drawings.
### Final Answer Completion
- The MAD of the data in the table is 10.
- If the additional drawings are Amy's, the MAD of the data set will decrease to 8.33.
- If they are Emily's, the MAD will increase to 12.22.
- The MAD of the new data set does depend on whether it was Amy or Emily who turned in the additional drawings.
- So, Greg is incorrect.