the table shows the growth, in centimeters, of plants in two different soils after two weeks.
soil type vs. plant growth
soil a: 3.1, 5.2, 4.6, 3.8, 6.0, 4.3, 3.7, 4.0, 3.5, 3.5
soil b: 2.9, 3.0, 5.0, 4.8, 4.5, 3.9, 3.9, 4.4, 5.5, 4.1
which statement explains which soil tended to have a greater growth?
- soil a had greater growth because its data have a greater standard deviation.
- soil a had greater growth because its data have a greater mean.
- soil b had greater growth because its data have a greater interquartile range.
- soil b had greater growth because its data have a greater median.

Question

Accepted Answer

To solve this, we calculate the mean, median, standard deviation, and interquartile range for both soils.

### Step 1: Calculate Mean for Soil A
Data for Soil A: $ 3.1, 5.2, 4.6, 3.8, 6.0, 4.3, 3.7, 4.0, 3.5, 3.5 $
Sum of values: $ 3.1 + 5.2 + 4.6 + 3.8 + 6.0 + 4.3 + 3.7 + 4.0 + 3.5 + 3.5 = 41.7 $
Number of values ($ n $): $ 10 $
Mean ($ \bar{x}_A $): $ \frac{41.7}{10} = 4.17 $

### Step 2: Calculate Mean for Soil B
Data for Soil B: $ 2.9, 3.0, 5.0, 4.8, 4.5, 3.9, 3.9, 4.4, 5.5, 4.1 $
Sum of values: $ 2.9 + 3.0 + 5.0 + 4.8 + 4.5 + 3.9 + 3.9 + 4.4 + 5.5 + 4.1 = 42.0 $
Number of values ($ n $): $ 10 $
Mean ($ \bar{x}_B $): $ \frac{42.0}{10} = 4.2 $ (Wait, correction: Wait, let's recalculate the sum for Soil B: $ 2.9 + 3.0 = 5.9 $; $ 5.9 + 5.0 = 10.9 $; $ 10.9 + 4.8 = 15.7 $; $ 15.7 + 4.5 = 20.2 $; $ 20.2 + 3.9 = 24.1 $; $ 24.1 + 3.9 = 28.0 $; $ 28.0 + 4.4 = 32.4 $; $ 32.4 + 5.5 = 37.9 $; $ 37.9 + 4.1 = 42.0 $. So mean is $ \frac{42.0}{10} = 4.2 $)

### Step 3: Calculate Median for Soil A
Sort Soil A data: $ 3.1, 3.5, 3.5, 3.7, 3.8, 4.0, 4.3, 4.6, 5.2, 6.0 $
Median (middle value for even $ n $): Average of 5th and 6th values. 5th value: $ 3.8 $, 6th value: $ 4.0 $
Median ($ M_A $): $ \frac{3.8 + 4.0}{2} = 3.9 $

### Step 4: Calculate Median for Soil B
Sort Soil B data: $ 2.9, 3.0, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5 $
Median (middle value for even $ n $): Average of 5th and 6th values. 5th value: $ 4.1 $, 6th value: $ 4.4 $
Median ($ M_B $): $ \frac{4.1 + 4.4}{2} = 4.25 $

### Step 5: Analyze Standard Deviation and Interquartile Range
- Standard Deviation: Measures spread, not central tendency (growth). So the first option is incorrect.
- Interquartile Range (IQR): Also measures spread, not central tendency. So the third option is incorrect.
- Median: Soil B's median ($ 4.25 $) is greater than Soil A's ($ 3.9 $), but let's check the mean again. Wait, earlier calculation for Soil A's mean: Wait, $ 3.1 + 5.2 = 8.3 $; $ 8.3 + 4.6 = 12.9 $; $ 12.9 + 3.8 = 16.7 $; $ 16.7 + 6.0 = 22.7 $; $ 22.7 + 4.3 = 27.0 $; $ 27.0 + 3.7 = 30.7 $; $ 30.7 + 4.0 = 34.7 $; $ 34.7 + 3.5 = 38.2 $; $ 38.2 + 3.5 = 41.7 $. So mean is $ 4.17 $. Soil B's mean: $ 42.0 / 10 = 4.2 $. Wait, but let's re - check the median calculation for Soil B. The sorted data for Soil B: $ 2.9, 3.0, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5 $. The 5th value is $ 4.1 $, 6th is $ 4.4 $, so median is $ (4.1 + 4.4)/2 = 4.25 $.

But wait, the options:
- Option 1: Standard deviation is about spread, not growth amount. Incorrect.
- Option 2: Soil A's mean is $ 4.17 $, Soil B's is $ 4.2 $. Wait, maybe I miscalculated Soil A's mean. Wait, $ 3.1+5.2 = 8.3 $; $ 8.3 + 4.6 = 12.9 $; $ 12.9+3.8 = 16.7 $; $ 16.7 + 6.0 = 22.7 $; $ 22.7+4.3 = 27.0 $; $ 27.0+3.7 = 30.7 $; $ 30.7+4.0 = 34.7 $; $ 34.7+3.5 = 38.2 $; $ 38.2+3.5 = 41.7 $. $ 41.7/10 = 4.17 $. Soil B: $ 2.9+3.0 = 5.9 $; $ 5.9+5.0 = 10.9 $; $ 10.9+4.8 = 15.7 $; $ 15.7+4.5 = 20.2 $; $ 20.2+3.9 = 24.1 $; $ 24.1+3.9 = 28.0 $; $ 28.0+4.4 = 32.4 $; $ 32.4+5.5 = 37.9 $; $ 37.9+4.1 = 42.0 $. $ 42.0/10 = 4.2 $. So Soil B's mean is slightly higher. But wait, the option "Soil B had greater growth because its data have a greater median" – let's check the median again. Soil A's sorted data: $ 3.1, 3.5, 3.5, 3.7, 3.8, 4.0, 4.3, 4.6, 5.2, 6.0 $. The median is the average of the 5th and 6th terms: $ (3.8 + 4.0)/2 = 3.9 $. Soil B's sorted data: $ 2.9, 3.0, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5 $. Median is $ (4.1 + 4.4)/2 = 4.25 $. So Soil B's median is greater. But wait, the option "Soil B had greater growth because its data have a greater median" – median is a measure of central tendency. Let's check the other options:

- Option 1: Standard deviation is spread, not growth. Incorrect.
- Option 2: Soil A's mean is $ 4.17 $, Soil B's is $ 4.2 $. Close, but not "greater" in a way that's a clear distinction? Wait, maybe I made a mistake in Soil A's data. Wait, Soil A has 10 data points: $ 3.1, 5.2, 4.6, 3.8, 6.0, 4.3, 3.7, 4.0, 3.5, 3.5 $. Let's sum again: $ 3.1 + 3.5 + 3.5 + 3.7 + 3.8 + 4.0 + 4.3 + 4.6 + 5.2 + 6.0 $. $ 3.1+3.5 = 6.6 $; $ 6.6+3.5 = 10.1 $; $ 10.1+3.7 = 13.8 $; $ 13.8+3.8 = 17.6 $; $ 17.6+4.0 = 21.6 $; $ 21.6+4.3 = 25.9 $; $ 25.9+4.6 = 30.5 $; $ 30.5+5.2 = 35.7 $; $ 35.7+6.0 = 41.7 $. Correct. Soil B: $ 2.9 + 3.0 + 3.9 + 3.9 + 4.1 + 4.4 + 4.5 + 4.8 + 5.0 + 5.5 $. Wait, no, the original Soil B data is $ 2.9, 3.0, 5.0, 4.8, 4.5, 3.9, 3.9, 4.4, 5.5, 4.1 $. Let's sort it correctly: $ 2.9, 3.0, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5 $. Yes, that's correct. So median is $ (4.1 + 4.4)/2 = 4.25 $.

Now, let's check the interquartile range (IQR) for Soil A:
- Q1 (25th percentile): For $ n = 10 $, the 25th percentile is the average of the 2nd and 3rd terms (since $ 0.25\times10 = 2.5 $). Sorted Soil A: $ 3.1, 3.5, 3.5, 3.7, 3.8, 4.0, 4.3, 4.6, 5.2, 6.0 $. 2nd term: $ 3.5 $, 3rd term: $ 3.5 $. Q1: $ (3.5 + 3.5)/2 = 3.5 $
- Q3 (75th percentile): Average of the 7th and 8th terms. 7th term: $ 4.3 $, 8th term: $ 4.6 $. Q3: $ (4.3 + 4.6)/2 = 4.45 $
- IQR for A: $ 4.45 - 3.5 = 0.95 $

For Soil B:
- Q1 (25th percentile): Average of 2nd and 3rd terms ($ 0.25\times10 = 2.5 $). Sorted Soil B: $ 2.9, 3.0, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5 $. 2nd term: $ 3.0 $, 3rd term: $ 3.9 $. Q1: $ (3.0 + 3.9)/2 = 3.45 $
- Q3 (75th percentile): Average of 7th and 8th terms. 7th term: $ 4.5 $, 8th term: $ 4.8 $. Q3: $ (4.5 + 4.8)/2 = 4.65 $
- IQR for B: $ 4.65 - 3.45 = 1.2 $. So Soil B has a greater IQR, but IQR is spread, not growth. So option 3 is incorrect.

Now, the mean: Soil A's mean is $ 4.17 $, Soil B's is $ 4.2 $. The difference is small, but let's check the median. Soil B's median ($ 4.25 $) is greater than Soil A's ($ 3.9 $). So the option "Soil B had greater growth because its data have a greater median" is correct? Wait, but let's re - evaluate the options:

- Option 1: Standard deviation is spread, not growth. Incorrect.
- Option 2: Soil A's mean is ~4.17, Soil B's is ~4.2. So Soil B's mean is slightly higher, but the option says "Soil A had greater growth because its data have a greater mean" – which is incorrect.
- Option 3: IQR is spread, not growth. Incorrect.
- Option 4: Soil B's median is greater than Soil A's, and median is a measure of central tendency (growth). So this is correct.

# Answer:
Soil B had greater growth because its data have a greater median. (The correct option is the fourth one: "Soil B had greater growth because its data have a greater median.")

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QUESTION IMAGE

Question

Explanation:

Step 1: Calculate Mean for Soil A

Step 2: Calculate Mean for Soil B

Step 3: Calculate Median for Soil A

Step 4: Calculate Median for Soil B

Step 5: Analyze Standard Deviation and Interquartile Range

Answer: