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?\nsoil a had greater growth because its data have a greater standard deviation.\nsoil a had greater growth because its data have a greater mean.\nsoil b had greater growth because its data have a greater interquartile range.\nsoil 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?\nsoil a had greater growth because its data have a greater standard deviation.\nsoil a had greater growth because its data have a greater mean.\nsoil b had greater growth because its data have a greater interquartile range.\nsoil b had greater growth because its data have a greater median.

Answer

Explanation:

Step1: Calculate mean of Soil A

$\text{Mean}_A=\frac{3.1 + 5.2+4.6+3.8+6.0+4.3+3.7+4.0+3.5+3.5}{10}=\frac{42.7}{10} = 4.27$

Step2: Calculate mean of Soil B

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

Step3: Analyze statements

  • Standard - deviation measures spread, not central tendency of growth.
  • Inter - quartile range measures spread, not central tendency of growth.
  • Since $\text{Mean}_B(4.28)>\text{Mean}_A(4.27)$, Soil B has a greater mean growth.
  • Calculate median of Soil A: Arrange data in ascending order: $3.1,3.5,3.5,3.7,3.8,4.0,4.3,4.6,5.2,6.0$. Median$=\frac{3.8 + 4.0}{2}=3.9$. Calculate median of Soil B: Arrange data in ascending order: $2.9,3.8,3.9,3.9,4.1,4.4,4.5,4.8,5.0,5.5$. Median$=\frac{4.1+4.4}{2}=4.25$. But mean is a better measure for comparing overall growth when data is not highly skewed.

Answer:

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