To determine which statements are true, we need to calculate the mean and standard deviation for both data sets.
Data Set 1: 5, 5, 6, 6, 7, 7
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Mean of Data Set 1: \[ \text{Mean} = \frac{5 + 5 + 6 + 6 + 7 + 7}{6} = \frac{36}{6} = 6 \]
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Standard Deviation of Data Set 1:
- First, find the variance: \[ \text{Variance} = \frac{(5-6)^2 + (5-6)^2 + (6-6)^2 + (6-6)^2 + (7-6)^2 + (7-6)^2}{6} \] \[ = \frac{1 + 1 + 0 + 0 + 1 + 1}{6} = \frac{4}{6} = \frac{2}{3} \]
- Standard deviation is the square root of variance: \[ \text{Standard Deviation} = \sqrt{\frac{2}{3}} \approx 0.816 \]
Data Set 2: 1, 3, 5, 7, 9, 11
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Mean of Data Set 2: \[ \text{Mean} = \frac{1 + 3 + 5 + 7 + 9 + 11}{6} = \frac{36}{6} = 6 \]
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Standard Deviation of Data Set 2:
- First, find the variance: \[ \text{Variance} = \frac{(1-6)^2 + (3-6)^2 + (5-6)^2 + (7-6)^2 + (9-6)^2 + (11-6)^2}{6} \] \[ = \frac{25 + 9 + 1 + 1 + 9 + 25}{6} = \frac{70}{6} \approx 11.67 \]
- Standard deviation is the square root of variance: \[ \text{Standard Deviation} = \sqrt{\frac{70}{6}} \approx 2.71 \]
Now, let’s evaluate the original statements:
- Data Set 2 has a smaller standard deviation than Data Set 1. - False (Data Set 2: ~2.71, Data Set 1: ~0.816)
- Data Set 2 has the same mean as Data Set 1. - True (Both means are 6)
- Data Set 2 has a larger standard deviation than Data Set 1. - True (Data Set 2: ~2.71, Data Set 1: ~0.816)
- Data Set 2 has a larger mean than Data Set 1. - False (Both means are 6)
Thus, the two true statements are:
- Data Set 2 has the same mean as Data Set 1.
- Data Set 2 has a larger standard deviation than Data Set 1.