To compare the variability of the two data sets using the standard deviation, we will calculate the standard deviation for each data set.
Data Set A: 4, 5, 6, 12, 13
-
Calculate the Mean (Average): \[ \text{Mean}_A = \frac{4 + 5 + 6 + 12 + 13}{5} = \frac{40}{5} = 8 \]
-
Calculate Each Deviation from the Mean:
- \(4 - 8 = -4\)
- \(5 - 8 = -3\)
- \(6 - 8 = -2\)
- \(12 - 8 = 4\)
- \(13 - 8 = 5\)
-
Square Each Deviation:
- \((-4)^2 = 16\)
- \((-3)^2 = 9\)
- \((-2)^2 = 4\)
- \(4^2 = 16\)
- \(5^2 = 25\)
-
Calculate the Variance: \[ \text{Variance}_A = \frac{16 + 9 + 4 + 16 + 25}{5} = \frac{70}{5} = 14 \]
-
Calculate the Standard Deviation: \[ \text{Standard Deviation}_A = \sqrt{14} \approx 3.74 \]
Data Set B: 4, 5, 7, 9, 10
-
Calculate the Mean (Average): \[ \text{Mean}_B = \frac{4 + 5 + 7 + 9 + 10}{5} = \frac{35}{5} = 7 \]
-
Calculate Each Deviation from the Mean:
- \(4 - 7 = -3\)
- \(5 - 7 = -2\)
- \(7 - 7 = 0\)
- \(9 - 7 = 2\)
- \(10 - 7 = 3\)
-
Square Each Deviation:
- \((-3)^2 = 9\)
- \((-2)^2 = 4\)
- \(0^2 = 0\)
- \(2^2 = 4\)
- \(3^2 = 9\)
-
Calculate the Variance: \[ \text{Variance}_B = \frac{9 + 4 + 0 + 4 + 9}{5} = \frac{26}{5} = 5.2 \]
-
Calculate the Standard Deviation: \[ \text{Standard Deviation}_B = \sqrt{5.2} \approx 2.28 \]
Summary of Results:
- Standard Deviation of Data Set A: \(\approx 3.74\)
- Standard Deviation of Data Set B: \(\approx 2.28\)
Conclusion:
The standard deviation of Data Set A (approximately 3.74) is greater than the standard deviation of Data Set B (approximately 2.28), indicating that Data Set A has greater variability compared to Data Set B.