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the table shows the growth, in centimeters, of plants in two different …

Question

the table shows the growth, in centimeters, of plants in two different soils after two weeks.
soil type vs. plant growth

soil a3.15.24.83.86.04.33.74.03.53.5
soil b2.93.85.04.84.53.93.94.45.54.1

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.

Explanation:

To determine which soil has greater growth, we analyze the mean (average) of each soil's plant growth data.

Step 1: Calculate the mean for Soil A

The data for Soil A is: \( 3.1, 5.2, 4.8, 3.8, 6.0, 4.3, 3.7, 4.0, 3.5, 3.5 \)
The sum of these values is:

$$ LATEXBLOCK0 $$

There are 10 data points, so the mean (\(\bar{x}_A\)) is:

$$ \bar{x}_A = \frac{41.9}{10} = 4.19 $$
Step 2: Calculate the mean for Soil B

The data for Soil B is: \( 2.9, 3.8, 5.0, 4.8, 4.5, 3.9, 3.9, 4.4, 5.5, 4.1 \)
The sum of these values is:

$$ LATEXBLOCK1 $$

There are 10 data points, so the mean (\(\bar{x}_B\)) is:

$$ \bar{x}_B = \frac{42.8}{10} = 4.28 $$
Step 3: Analyze the options
  • Option 1: Standard deviation measures spread, not central tendency (growth). Eliminate.
  • Option 2: Soil A's mean is \( 4.19 \), Soil B's is \( 4.28 \). Wait, no—wait, recalculate Soil A's sum. Wait, let's recheck Soil A's data: \( 3.1, 5.2, 4.8, 3.8, 6.0, 4.3, 3.7, 4.0, 3.5, 3.5 \). Let's sum again:

\( 3.1 + 5.2 = 8.3 \); \( 8.3 + 4.8 = 13.1 \); \( 13.1 + 3.8 = 16.9 \); \( 16.9 + 6.0 = 22.9 \); \( 22.9 + 4.3 = 27.2 \); \( 27.2 + 3.7 = 30.9 \); \( 30.9 + 4.0 = 34.9 \); \( 34.9 + 3.5 = 38.4 \); \( 38.4 + 3.5 = 41.9 \). Mean: \( 41.9 / 10 = 4.19 \). Soil B: \( 2.9 + 3.8 = 6.7 \); \( 6.7 + 5.0 = 11.7 \); \( 11.7 + 4.8 = 16.5 \); \( 16.5 + 4.5 = 21.0 \); \( 21.0 + 3.9 = 24.9 \); \( 24.9 + 3.9 = 28.8 \); \( 28.8 + 4.4 = 33.2 \); \( 33.2 + 5.5 = 38.7 \); \( 38.7 + 4.1 = 42.8 \). Mean: \( 42.8 / 10 = 4.28 \). Wait, maybe I made a mistake earlier. Wait, no—wait, the problem's Soil B data: let's check again. Wait, the original table: Soil B has 10 data points? Wait, the user's table: Soil A: 10 numbers, Soil B: 10 numbers? Wait, the user's table:

Soil A: 3.1, 5.2, 4.8, 3.8, 6.0, 4.3, 3.7, 4.0, 3.5, 3.5 (10)

Soil B: 2.9, 3.8, 5.0, 4.8, 4.5, 3.9, 3.9, 4.4, 5.5, 4.1 (10). Wait, maybe my calculation for Soil B was wrong. Let's sum Soil B again:

2.9 + 3.8 = 6.7

6.7 + 5.0 = 11.7

11.7 + 4.8 = 16.5

16.5 + 4.5 = 21.0

21.0 + 3.9 = 24.9

24.9 + 3.9 = 28.8

28.8 + 4.4 = 33.2

33.2 + 5.5 = 38.7

38.7 + 4.1 = 42.8. Yes, that's correct. So mean of B is 4.28, mean of A is 4.19. Wait, but the options: let's check the median.

Step 3 (Alternative: Median)
  • Median of Soil A: First, sort the data: \( 3.1, 3.5, 3.5, 3.7, 3.8, 4.0, 4.3, 4.8, 5.2, 6.0 \). There are 10 data points, so the median is the average of the 5th and 6th values: \( \frac{3.8 + 4.0}{2} = 3.9 \)
  • Median of Soil B: Sort the data: \( 2.9, 3.8, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5 \). Wait, no—wait, the data is \( 2.9, 3.8, 5.0, 4.8, 4.5, 3.9, 3.9, 4.4, 5.5, 4.1 \). Sorting: \( 2.9, 3.8, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5 \). Median is average of 5th and 6th: \( \frac{4.1 + 4.4}{2} = 4.25 \)

Now, let's re-express the options:

  • Option 1: Standard deviation (spread) doesn't indicate growth (central tendency). Eliminate.
  • Option 2: Soil A's mean is 4.19, Soil B's is 4.28. Wait, maybe I miscalculated Soil A's sum. Wait, Soil A data: 3.1, 5.2, 4.8, 3.8, 6.0, 4.3, 3.7, 4.0, 3.5, 3.5. Let's su…
Brief Explanations

To determine which soil has greater growth, we analyze central tendency (mean/median) as they reflect typical growth.

  • Mean: Soil A’s mean = \( \frac{41.9}{10} = 4.19 \); Soil B’s mean = \( \frac{42.8}{10} = 4.28 \).
  • Median: Soil A (sorted: \( 3.1, 3.5, 3.5, 3.7, 3.8, 4.0, 4.3, 4.8, 5.2, 6.0 \)) has median \( \frac{3.8 + 4.0}{2} = 3.9 \). Soil B (sorted: \( 2.9, 3.8, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5 \)) has median \( \frac{4.1 + 4.4}{2} = 4.25 \).

Standard deviation (spread) and interquartile range (spread) do not indicate typical growth. Soil B’s median (and mean) is greater, so it had greater growth.

Answer:

Soil B had greater growth because its data have a greater median.