Question

Based on the two data sets given below, which of the following statements are true? Select two that apply. Data Set 1: 5, 5, 6, 6, 7, 7 Data Set 2: 1, 3, 5, 7, 9, 11 (1 point) Responses Data Set 2 has a larger mean than Data Set 1. Data Set 2 has a larger mean than Data Set 1. Data Set 2 has a smaller standard deviation than Data Set 1. Data Set 2 has a smaller standard deviation than Data Set 1. Data Set 2 has the same mean as Data Set 1. Data Set 2 has the same mean as Data Set 1. Data Set 2 has a larger standard deviation than Data Set 1.

Answer 1

To analyze the two datasets, we will calculate the mean and standard deviation for both Data Set 1 and Data Set 2.

**Data Set 1: 5, 5, 6, 6, 7, 7**

1. **Mean of Data Set 1:**
\[
\text{Mean} = \frac{(5 + 5 + 6 + 6 + 7 + 7)}{6} = \frac{36}{6} = 6
\]

2. **Standard Deviation of Data Set 1:**
- First, find the variance:
\[
\text{Variance} = \frac{\sum (x_i - \text{mean})^2}{N} = \frac{(5-6)^2 + (5-6)^2 + (6-6)^2 + (6-6)^2 + (7-6)^2 + (7-6)^2}{6}
\]
\[
= \frac{(1 + 1 + 0 + 0 + 1 + 1)}{6} = \frac{4}{6} \approx 0.67
\]
- Standard deviation:
\[
\text{SD} = \sqrt{0.67} \approx 0.82
\]

**Data Set 2: 1, 3, 5, 7, 9, 11**

1. **Mean of Data Set 2:**
\[
\text{Mean} = \frac{(1 + 3 + 5 + 7 + 9 + 11)}{6} = \frac{36}{6} = 6
\]

2. **Standard Deviation of Data Set 2:**
- First, calculate the variance:
\[
\text{Variance} = \frac{\sum (x_i - \text{mean})^2}{N} = \frac{(1-6)^2 + (3-6)^2 + (5-6)^2 + (7-6)^2 + (9-6)^2 + (11-6)^2}{6}
\]
\[
= \frac{(25 + 9 + 1 + 1 + 9 + 25)}{6} = \frac{70}{6} \approx 11.67
\]
- Standard deviation:
\[
\text{SD} = \sqrt{11.67} \approx 3.42
\]

Now, summarizing our findings for the two data sets:

- **Mean:**
- Data Set 1 Mean = 6
- Data Set 2 Mean = 6

- **Standard Deviation:**
- Data Set 1 SD ≈ 0.82
- Data Set 2 SD ≈ 3.42

**Result:**

- Data Set 2 has the same mean as Data Set 1.
- Data Set 2 has a larger standard deviation than Data Set 1.

Thus, the two true statements are:

1. Data Set 2 has the same mean as Data Set 1.
2. Data Set 2 has a larger standard deviation than Data Set 1.