Question

if a new data point at 12 is added to the graph, which will be true?

• the median will increase, and the mean will stay the same.
• the mean will increase more than the median, but both will increase.
• the median will increase more than the mean, but both will increase.
• the mean will increase, and the median will stay the same.

Explanation:

<pre_analysis>
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<reasoning>

List the original data points

The dot plot shows the following data values:

• One dot at 1
• Three dots at 2
• One dot at 3
• Three dots at 4
• One dot at 5

The dataset is:
\[
\{1, 2, 2, 2, 3, 4, 4, 4, 5\}
\]
The number of data points is \(n = 9\).

Calculate initial mean and median

The sum of the data is:
\[
1 + 2\times3 + 3 + 4\times3 + 5 = 1 + 6 + 3 + 12 + 5 = 27
\]
The initial mean is:
\[
\mu_{\text{initial}} = \frac{27}{9} = 3
\]
Since \(n = 9\) is odd, the median is the 5th value in the ordered list:
\[
\text{Median}_{\text{initial}} = 3
\]

Calculate new mean and median with the outlier

Adding a new data point at 12, the new dataset is:
\[
\{1, 2, 2, 2, 3, 4, 4, 4, 5, 12\}
\]
The new number of data points is \(n = 10\).
The new sum is:
\[
27 + 12 = 39
\]
The new mean is:
\[
\mu_{\text{new}} = \frac{39}{10} = 3.9
\]
Since \(n = 10\) is even, the median is the average of the 5th and 6th values (3 and 4):
\[
\text{Median}_{\text{new}} = \frac{3 + 4}{2} = 3.5
\]

Compare the changes

• The mean increased from 3 to 3.9 (an increase of 0.9).
• The median increased from 3 to 3.5 (an increase of 0.5).
• Both measures increased, but the mean increased more than the median because 12 is an extreme outlier.

</reasoning>

<answer>
<mcq-option>The median will increase, and the mean will stay the same.</mcq-option>
<mcq-correct>The mean will increase more than the median, but both will increase.</mcq-correct>
<mcq-option>The median will increase more than the mean, but both will increase.</mcq-option>
<mcq-option>The mean will increase, and the median will stay the same.</mcq-option>
</answer>

<post_analysis>
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"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Statistics",
"Outlier Effects on Mean"
]
}
</post_analysis>

Answer:

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</pre_analysis>

<reasoning>

List the original data points

The dot plot shows the following data values:

• One dot at 1
• Three dots at 2
• One dot at 3
• Three dots at 4
• One dot at 5

The dataset is:
\[
\{1, 2, 2, 2, 3, 4, 4, 4, 5\}
\]
The number of data points is \(n = 9\).

Calculate initial mean and median

The sum of the data is:
\[
1 + 2\times3 + 3 + 4\times3 + 5 = 1 + 6 + 3 + 12 + 5 = 27
\]
The initial mean is:
\[
\mu_{\text{initial}} = \frac{27}{9} = 3
\]
Since \(n = 9\) is odd, the median is the 5th value in the ordered list:
\[
\text{Median}_{\text{initial}} = 3
\]

Calculate new mean and median with the outlier

Adding a new data point at 12, the new dataset is:
\[
\{1, 2, 2, 2, 3, 4, 4, 4, 5, 12\}
\]
The new number of data points is \(n = 10\).
The new sum is:
\[
27 + 12 = 39
\]
The new mean is:
\[
\mu_{\text{new}} = \frac{39}{10} = 3.9
\]
Since \(n = 10\) is even, the median is the average of the 5th and 6th values (3 and 4):
\[
\text{Median}_{\text{new}} = \frac{3 + 4}{2} = 3.5
\]

Compare the changes

• The mean increased from 3 to 3.9 (an increase of 0.9).
• The median increased from 3 to 3.5 (an increase of 0.5).
• Both measures increased, but the mean increased more than the median because 12 is an extreme outlier.

</reasoning>

<answer>
<mcq-option>The median will increase, and the mean will stay the same.</mcq-option>
<mcq-correct>The mean will increase more than the median, but both will increase.</mcq-correct>
<mcq-option>The median will increase more than the mean, but both will increase.</mcq-option>
<mcq-option>The mean will increase, and the median will stay the same.</mcq-option>
</answer>

<post_analysis>
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"subject": "Mathematics",
"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Statistics",
"Outlier Effects on Mean"
]
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</post_analysis>