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

the table shows the growth, in centimeters, of plants in two different soils after two weeks.\nsoil type vs. plant growth\n|soil a|3.1|5.2|4.6|3.8|6.0|4.3|3.7|4.0|3.5|3.5|\n|soil b|2.9|3.8|5.0|4.8|4.5|3.9|3.9|4.4|5.5|4.1|\nwhich statement explains which soil tended to have a greater growth?\no soil a had greater growth because its data have a greater standard deviation.\no soil a had greater growth because its data have a greater mean.\no soil b had greater growth because its data have a greater interquartile range.\no 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.\nsoil type vs. plant growth\n|soil a|3.1|5.2|4.6|3.8|6.0|4.3|3.7|4.0|3.5|3.5|\n|soil b|2.9|3.8|5.0|4.8|4.5|3.9|3.9|4.4|5.5|4.1|\nwhich statement explains which soil tended to have a greater growth?\no soil a had greater growth because its data have a greater standard deviation.\no soil a had greater growth because its data have a greater mean.\no soil b had greater growth because its data have a greater interquartile range.\no soil b had greater growth because its data have a greater median.

Answer

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 statements

Standard - deviation measures spread, not central - tendency for growth comparison. Inter - quartile range also measures spread. Mean and median are measures of central tendency. Since $\bar{x}_B>\bar{x}_A$, Soil B has a greater mean.

Answer:

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