To evaluate the statements, we first need to calculate the mean and standard deviation for both data sets.
Data set A: 4, 5, 6, 12, 13
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Mean of Data set A: \[ \text{Mean} = \frac{4 + 5 + 6 + 12 + 13}{5} = \frac{40}{5} = 8 \]
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Standard Deviation of Data set A:
- First, find the variance: \[ \text{Variance} = \frac{(4-8)^2 + (5-8)^2 + (6-8)^2 + (12-8)^2 + (13-8)^2}{5} \] \[ = \frac{(-4)^2 + (-3)^2 + (-2)^2 + (4)^2 + (5)^2}{5} = \frac{16 + 9 + 4 + 16 + 25}{5} = \frac{70}{5} = 14 \]
- Standard Deviation: \[ \text{Standard Deviation} = \sqrt{14} \approx 3.74 \]
Data set B: 4, 5, 7, 9, 10
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Mean of Data set B: \[ \text{Mean} = \frac{4 + 5 + 7 + 9 + 10}{5} = \frac{35}{5} = 7 \]
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Standard Deviation of Data set B:
- First, find the variance: \[ \text{Variance} = \frac{(4-7)^2 + (5-7)^2 + (7-7)^2 + (9-7)^2 + (10-7)^2}{5} \] \[ = \frac{(-3)^2 + (-2)^2 + (0)^2 + (2)^2 + (3)^2}{5} = \frac{9 + 4 + 0 + 4 + 9}{5} = \frac{26}{5} = 5.2 \]
- Standard Deviation: \[ \text{Standard Deviation} = \sqrt{5.2} \approx 2.28 \]
Now, we have:
- Mean of Data set A: 8
- Mean of Data set B: 7
- Standard Deviation of Data set A: \(\sqrt{14} \approx 3.74\)
- Standard Deviation of Data set B: \(\sqrt{5.2} \approx 2.28\)
Now we evaluate the statements:
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Data set A has the same mean as data set B.
False (Mean A is 8, Mean B is 7) -
Data set A has a smaller standard deviation than data set B.
False (Standard Deviation A is approximately 3.74, Standard Deviation B is approximately 2.28) -
Data set A has a larger standard deviation than data set B.
True (3.74 > 2.28) -
Data set A has a larger mean than data set B.
True (Mean A is 8, Mean B is 7)
So the two true statements are:
- Data set A has a larger standard deviation than data set B.
- Data set A has a larger mean than data set B.