Use the standard deviation to compare the variability of the datasets. Round to the nearest hundredths.

Question

Dataset A: 11, 7, 12, 8, 9, 5

Dataset B: 10, 8, 10, 8, 8, 4

(1 point)
Responses

1.The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.58. Dataset A and Dataset B have the same standard deviation, so the variability is the same for both datasets.
The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.58. Dataset A and Dataset B have the same standard deviation, so the variability is the same for both datasets.

2.The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.19. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.
The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.19. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.

3.The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.19. The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability.
The standard deviation of Dataset A is 2.58. The standard deviation of Dataset B is 2.19. The standard deviation of Dataset A is larger than that of Dataset B, so Dataset A has more variability.

4.The standard deviation of Dataset A is 2.19. The standard deviation of Dataset B is 2.58. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.

answer using 1-4

Answer 1

To find the standard deviation for both datasets, we can follow these steps:

1. Calculate the mean (average) of each dataset.
2. Calculate the variance by finding the average of the squared differences from the mean.
3. Take the square root of the variance to find the standard deviation.

Let's calculate the standard deviations for both datasets.

Dataset A: 11, 7, 12, 8, 9, 5

1. Mean: $$\text{Mean}=\frac{11+7+12+8+9+5}{6}=\frac{52}{6}\approx 8.67$$

2. Variance: $$\begin{array}{rlrlr}\text{Variance}& =\frac{(11-8.67{)}^2+(7-8.67{)}^2+(12-8.67{)}^2+(8-8.67{)}^2+(9-8.67{)}^2+(5-8.67{)}^2}{6}& =\frac{(2.33{)}^2+(-1.67{)}^2+(3.33{)}^2+(-0.67{)}^2+(0.33{)}^2+(-3.67{)}^2}{6}& =\frac{5.4289+2.7889+11.0889+0.4489+0.1089+13.4689}{6}& =\frac{33.4134}{6}\approx 5.57\end{array}$$

3. Standard Deviation: $$\text{Standard Deviation}=\sqrt{5.57}\approx 2.36$$

Dataset B: 10, 8, 10, 8, 8, 4

1. Mean: $$\text{Mean}=\frac{10+8+10+8+8+4}{6}=\frac{58}{6}\approx 9.67$$

2. Variance: $$\begin{array}{rlrlr}\text{Variance}& =\frac{(10-9.67{)}^2+(8-9.67{)}^2+(10-9.67{)}^2+(8-9.67{)}^2+(8-9.67{)}^2+(4-9.67{)}^2}{6}& =\frac{(0.33{)}^2+(-1.67{)}^2+(0.33{)}^2+(-1.67{)}^2+(-1.67{)}^2+(-5.67{)}^2}{6}& =\frac{0.1089+2.7889+0.1089+2.7889+2.7889+32.1489}{6}& =\frac{41.7334}{6}\approx 6.95\end{array}$$

3. Standard Deviation: $$\text{Standard Deviation}=\sqrt{6.95}\approx 2.64$$

Summary of Results

• Standard deviation of Dataset A: 2.36
• Standard deviation of Dataset B: 2.64

Comparing both, the standard deviation of Dataset B is larger than that of Dataset A, which indicates that Dataset B has more variability.

Therefore, the correct answer is: 2. The standard deviation of Dataset A is 2.36. The standard deviation of Dataset B is 2.64. The standard deviation of Dataset B is larger than that of Dataset A, so Dataset B has more variability.