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

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.

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.

Answer

Answer:

B. Soil A had greater growth because its data have a greater mean.

Explanation:

Step1: Calculate mean of Soil A

$\bar{x}_A=\frac{3.1 + 5.2+4.6+3.8+6.0+4.3+3.7+4.0+3.5+3.5}{10}=\frac{41.7}{10} = 4.17$

Step2: Calculate mean of Soil B

$\bar{x}_B=\frac{2.9+3.8+5.0+4.8+4.5+3.9+3.9+4.4+5.5+4.1}{10}=\frac{42.8}{10}=4.28$

Step3: Analyze other measures

Standard - deviation measures spread, not central - tendency. Inter - quartile range also measures spread. Median of Soil A: Arrange data: $3.1,3.5,3.5,3.7,3.8,4.0,4.3,4.6,5.2,6.0$. Median$=\frac{3.8 + 4.0}{2}=3.9$. Median of Soil B: Arrange data: $2.9,3.8,3.9,3.9,4.1,4.4,4.5,4.8,5.0,5.5$. Median$=\frac{4.1+4.4}{2}=4.25$. But mean is a better measure for overall growth tendency. Since $4.28>4.17$, Soil B has a greater mean and thus greater overall growth tendency.