QUESTION IMAGE
Question
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 3.8 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.
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 \( = 3.1 + 5.2 + 4.6 + 3.8 + 6.0 + 4.3 + 3.7 + 4.0 + 3.5 + 3.5 = 41.7 \)
Mean \( (\mu_A) = \frac{41.7}{10} = 4.17 \)
Step 2: Calculate Mean for Soil B
Data for Soil B: \( 2.9, 3.8, 5.0, 4.8, 4.5, 3.9, 3.9, 4.4, 5.5, 4.1 \)
Sum \( = 2.9 + 3.8 + 5.0 + 4.8 + 4.5 + 3.9 + 3.9 + 4.4 + 5.5 + 4.1 = 42.8 \)
Mean \( (\mu_B) = \frac{42.8}{10} = 4.28 \)? Wait, no, wait—wait, let's re - calculate the sum for Soil B:
\( 2.9+3.8 = 6.7\); \( 6.7 + 5.0=11.7\); \( 11.7+4.8 = 16.5\); \( 16.5+4.5 = 21\); \( 21+3.9 = 24.9\); \( 24.9+3.9 = 28.8\); \( 28.8+4.4 = 33.2\); \( 33.2+5.5 = 38.7\); \( 38.7+4.1 = 42.8\). Mean is \( 42.8\div10 = 4.28\)? Wait, no, earlier for Soil A, sum was 41.7, mean 4.17. Wait, maybe I made a mistake. Wait, let's re - check Soil A's sum:
Soil A: \( 3.1+5.2 = 8.3\); \( 8.3+4.6 = 12.9\); \( 12.9+3.8 = 16.7\); \( 16.7+6.0 = 22.7\); \( 22.7+4.3 = 27\); \( 27+3.7 = 30.7\); \( 30.7+4.0 = 34.7\); \( 34.7+3.5 = 38.2\); \( 38.2+3.5 = 41.7\). Mean is \( 41.7/10 = 4.17\)
Soil B: Let's sort the data for median and IQR.
Soil A sorted: \( 3.1, 3.5, 3.5, 3.7, 3.8, 4.0, 4.3, 4.6, 5.2, 6.0\)
Median of Soil A: Since there are 10 data points, median is average of 5th and 6th terms. 5th term = 3.8, 6th term = 4.0. Median \( = \frac{3.8 + 4.0}{2}=3.9\)
Soil B sorted: \( 2.9, 3.8, 3.9, 3.9, 4.1, 4.4, 4.5, 4.8, 5.0, 5.5\)
Median of Soil B: average of 5th and 6th terms. 5th term = 4.1, 6th term = 4.4. Median \(=\frac{4.1 + 4.4}{2}=4.25\)
Now, standard deviation: Standard deviation measures spread, not central tendency (growth). Interquartile range also measures spread. To determine which soil has greater growth, we use measures of central tendency (mean, median).
Mean of Soil A: 4.17, Mean of Soil B: Let's recalculate Soil B's sum. \( 2.9+3.8 = 6.7\); \( 6.7+3.9 = 10.6\); \( 10.6+3.9 = 14.5\); \( 14.5+4.1 = 18.6\); \( 18.6+4.4 = 23\); \( 23+4.5 = 27.5\); \( 27.5+4.8 = 32.3\); \( 32.3+5.0 = 37.3\); \( 37.3+5.5 = 42.8\). Mean is \( 42.8\div10 = 4.28\). Wait, but earlier when we calculated median, Soil B's median is 4.25, Soil A's median is 3.9. Wait, but the options:
Option 1: Standard deviation is about spread, not growth. So wrong.
Option 2: Wait, maybe I miscalculated Soil A's mean. Wait, let's re - sum Soil A:
3.1 + 5.2 = 8.3; +4.6 = 12.9; +3.8 = 16.7; +6.0 = 22.7; +4.3 = 27.0; +3.7 = 30.7; +4.0 = 34.7; +3.5 = 38.2; +3.5 = 41.7. Mean 4.17.
Soil B: 2.9+3.8 = 6.7; +5.0 = 11.7; +4.8 = 16.5; +4.5 = 21.0; +3.9 = 24.9; +3.9 = 28.8; +4.4 = 33.2; +5.5 = 38.7; +4.1 = 42.8. Mean 4.28. Wait, but the option says "Soil A had greater growth because its data have a greater mean"—but Soil B's mean is higher. Wait, maybe I made a mistake. Wait, let's check the original data again.
Wait, the Soil A data: 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 = 6.6; +3.5 = 10.1; +3.7 = 13.8; +3.8 = 17.6; +4.0 = 21.6; +4.3 = 25.9; +4.6 = 30.5; +5.2 = 35.7; +6.0 = 41.7. Yes, mean 4.17.
Soil B data: 2.9, 3.8, 5.0, 4.8, 4.5, 3.9, 3.9, 4.4, 5.5, 4.1. Let's sum:
2.9+3.8 = 6.7; +3.9 = 10.6; +3.9 = 14.5; +4.1 = 18.6; +4.4 = 23.0; +4.5 = 27.5; +4.8 = 32.3; +5.0 = 37.3; +5.5 = 42.8. Mean 4.28. Wait, but the option says "Soil A had greater growth because its data have a greater mean"—but that's not true. Wait, maybe I sorted the data wrong. Wait, no—wait, maybe the question's options are based on correct calculations. Wait…
To determine which soil has greater growth, we analyze measures of central tendency (mean, median) and spread (standard deviation, interquartile range). Spread - related measures (standard deviation, interquartile range) do not indicate greater growth. For central tendency:
- Mean: Soil A mean = \( \frac{41.7}{10}=4.17 \), Soil B mean = \( \frac{42.8}{10}=4.28 \) (Soil B has a greater mean, but the option about Soil A’s mean is incorrect).
- Median: Soil A median (average of 5th and 6th sorted terms) = \( \frac{3.8 + 4.0}{2}=3.9 \); Soil B median (average of 5th and 6th sorted terms) = \( \frac{4.1 + 4.4}{2}=4.25 \). Soil B’s median is greater.
- Spread measures: Standard deviation and interquartile range (IQR) measure variability, not growth. Soil A has a greater IQR, and spread does not indicate greater growth.
Thus, the correct statement is that Soil B had greater growth because its data have a greater median.
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Soil B had greater growth because its data have a greater median.