Calculating with Data

11 questions

Question 1Question →

A researcher conducted three trials to measure the mass of dissolved solute in 100 mL100\text{ mL} of water at 25∘C25^\circ\text{C}:

TrialMass of Dissolved Solute (g)
Trial 112.0
Trial 214.0
Trial 316.0

True or False: The average mass of dissolved solute across all three trials is 14.0 g14.0\text{ g}.

Show answer & explanation

Answer: True

Answer

True. The calculated mean of the three experimental trials is exactly 14.0 g.
Adding the three trial measurements (12.0 g+14.0 g+16.0 g=42.0 g12.0\text{ g} + 14.0\text{ g} + 16.0\text{ g} = 42.0\text{ g}) and dividing by the total number of trials (33) gives 42.0 g3=14.0 g\frac{42.0\text{ g}}{3} = 14.0\text{ g}. Thus, the statement is true.

Step-by-Step Solution

1
Retrieve the mass values for Trial 1, Trial 2, and Trial 3 from the data table.
Trial 1 = 12.0 g12.0\text{ g}, Trial 2 = 14.0 g14.0\text{ g}, Trial 3 = 16.0 g16.0\text{ g}.
Identify all data points needed to calculate the arithmetic mean.
2
Sum the retrieved mass values.
12.0 g+14.0 g+16.0 g=42.0 g12.0\text{ g} + 14.0\text{ g} + 16.0\text{ g} = 42.0\text{ g}.
Calculate the total mass recorded across all three trials.
3
Divide the total mass by the number of trials (33).
42.0 g3=14.0 g\frac{42.0\text{ g}}{3} = 14.0\text{ g}.
Obtain the average value per trial.

Key Concept

Calculating the average of data points collected from experimental trials
Question 2Question →

A microbiologist measured the bacterial growth rate (in cells/mL⋅hr\text{cells/mL}\cdot\text{hr}) of a newly isolated strain across three independent trials under four different incubator temperatures, as presented in the table below:

Temperature (∘C^\circ\text{C})Trial 1 (cells/mL⋅hr\text{cells/mL}\cdot\text{hr})Trial 2 (cells/mL⋅hr\text{cells/mL}\cdot\text{hr})Trial 3 (cells/mL⋅hr\text{cells/mL}\cdot\text{hr})
25140155125
30280310250
35420460380
40210240180

True or False: The mean growth rate at 35∘C35^\circ\text{C} exceeds the mean growth rate at 30∘C30^\circ\text{C} by more than 130 cells/mL⋅hr130\text{ cells/mL}\cdot\text{hr}.

Show answer & explanation

Answer: True

Answer

The statement is True because the mean growth rate at 35∘C35^\circ\text{C} (420 cells/mL⋅hr420\text{ cells/mL}\cdot\text{hr}) exceeds the mean growth rate at 30∘C30^\circ\text{C} (280 cells/mL⋅hr280\text{ cells/mL}\cdot\text{hr}) by 140 cells/mL⋅hr140\text{ cells/mL}\cdot\text{hr}, which is greater than 130 cells/mL⋅hr130\text{ cells/mL}\cdot\text{hr}.
The mean growth rate at 35∘C35^\circ\text{C} is 420 cells/mL⋅hr420\text{ cells/mL}\cdot\text{hr} and at 30∘C30^\circ\text{C} is 280 cells/mL⋅hr280\text{ cells/mL}\cdot\text{hr}. The difference is 140 cells/mL⋅hr140\text{ cells/mL}\cdot\text{hr}, which is strictly greater than 130 cells/mL⋅hr130\text{ cells/mL}\cdot\text{hr}.

Step-by-Step Solution

1
Calculate the average growth rate for 35∘C35^\circ\text{C} across all three trials.
Mean35=420+460+3803=12603=420 cells/mL⋅hr\text{Mean}_{35} = \frac{420 + 460 + 380}{3} = \frac{1260}{3} = 420\text{ cells/mL}\cdot\text{hr}
Determining the representative data point for the 35∘C35^\circ\text{C} temperature condition requires finding the mean.
2
Calculate the average growth rate for 30∘C30^\circ\text{C} across all three trials.
Mean30=280+310+2503=8403=280 cells/mL⋅hr\text{Mean}_{30} = \frac{280 + 310 + 250}{3} = \frac{840}{3} = 280\text{ cells/mL}\cdot\text{hr}
Determining the representative data point for the 30∘C30^\circ\text{C} temperature condition requires finding the mean.
3
Subtract the mean rate at 30∘C30^\circ\text{C} from the mean rate at 35∘C35^\circ\text{C}.
420−280=140 cells/mL⋅hr420 - 280 = 140\text{ cells/mL}\cdot\text{hr}
Finding the increase in average growth rate between the two temperature conditions.
4
Compare the calculated difference (140 cells/mL⋅hr140\text{ cells/mL}\cdot\text{hr}) to the value given in the statement (130 cells/mL⋅hr130\text{ cells/mL}\cdot\text{hr}).
140>130140 > 130, so the statement is True.
Verifying whether the statement's inequality holds.

Key Concept

Calculating multi-trial averages and evaluating differences between data sets.
Question 3Question →

Soil ecologists evaluated nitrogen mineralization by measuring nitrate production rates (in mg/kg/day\text{mg/kg/day}) across 4 depth zones in two distinct forest management plots (Plot X and Plot Y). The results are summarized in Table 1 below.

Depth ZonePlot X Nitrate Rate (mg/kg/day\text{mg/kg/day})Plot Y Nitrate Rate (mg/kg/day\text{mg/kg/day})
0–10 cm5.09.0
10–20 cm3.56.5
20–30 cm2.04.0
30–40 cm1.52.5

Based on Table 1, if a composite sample is created using equal masses of soil from all 4 depth zones (0–40 cm), by how much does the average nitrate production rate of Plot Y exceed the average nitrate production rate of Plot X?

Show answer & explanation

Answer: 2.5 mg/kg/day2.5\text{ mg/kg/day}

Answer

The average nitrate production rate of Plot Y exceeds that of Plot X by 2.5 mg/kg/day2.5\text{ mg/kg/day}.
The composite average rate for Plot Y is calculated by summing its four depth zone values (9.0+6.5+4.0+2.5=22.09.0 + 6.5 + 4.0 + 2.5 = 22.0) and dividing by 44, yielding 5.5 mg/kg/day5.5\text{ mg/kg/day}. The composite average rate for Plot X is calculated similarly (5.0+3.5+2.0+1.5=12.05.0 + 3.5 + 2.0 + 1.5 = 12.0) and divided by 44, yielding 3.0 mg/kg/day3.0\text{ mg/kg/day}. The difference between these two averages is 5.5−3.0=2.5 mg/kg/day5.5 - 3.0 = 2.5\text{ mg/kg/day}.

Step-by-Step Solution

1
Calculate the average nitrate production rate for Plot Y across the 4 depth zones.
Average for Plot Y = 9.0+6.5+4.0+2.54=22.04=5.5 mg/kg/day\frac{9.0 + 6.5 + 4.0 + 2.5}{4} = \frac{22.0}{4} = 5.5\text{ mg/kg/day}.
An equal-mass composite sample over 0–40 cm requires finding the arithmetic mean of all 4 depth intervals.
2
Calculate the average nitrate production rate for Plot X across the 4 depth zones.
Average for Plot X = 5.0+3.5+2.0+1.54=12.04=3.0 mg/kg/day\frac{5.0 + 3.5 + 2.0 + 1.5}{4} = \frac{12.0}{4} = 3.0\text{ mg/kg/day}.
The same averaging procedure must be applied to Plot X to determine its overall composite rate.
3
Subtract the average rate of Plot X from the average rate of Plot Y.
5.5 mg/kg/day−3.0 mg/kg/day=2.5 mg/kg/day5.5\text{ mg/kg/day} - 3.0\text{ mg/kg/day} = 2.5\text{ mg/kg/day}.
The question asks by how much Plot Y's composite average exceeds Plot X's composite average.

Key Concept

Calculating mean values from multi-source tabular data and determining net differences between treatments.
Question 4Question →

Biochemists measured the rate of glucose consumption (in mM/min\text{mM/min}) by a newly isolated bacterial strain across 3 temperature treatments (25∘C25^\circ\text{C}, 35∘C35^\circ\text{C}, and 45∘C45^\circ\text{C}). Each temperature condition was evaluated in 3 separate trials. The results are recorded in the table below:

Temperature (∘C^\circ\text{C})Trial 1 (mM/min\text{mM/min})Trial 2 (mM/min\text{mM/min})Trial 3 (mM/min\text{mM/min})
25251.21.21.51.51.81.8
35354.04.04.64.64.64.6
45452.12.12.52.52.02.0

Based on the data provided, what is the average rate of glucose consumption, in mM/min\text{mM/min}, for the 35∘C35^\circ\text{C} treatment across the 3 trials?

Show answer & explanation

Answer: 4.4

Answer

The average rate of glucose consumption at 35∘C35^\circ\text{C} across the 3 trials is 4.4 mM/min4.4\text{ mM/min}.
To find the average glucose consumption rate at 35∘C35^\circ\text{C}, locate the row for 35∘C35^\circ\text{C} in the table and sum the values across the 3 trials (4.0+4.6+4.6=13.2 mM/min4.0 + 4.6 + 4.6 = 13.2\text{ mM/min}). Then divide by the number of trials (33) to get 13.23=4.4 mM/min\frac{13.2}{3} = 4.4\text{ mM/min}.

Step-by-Step Solution

1
Extract data points for the 35∘C35^\circ\text{C} row
Trial 1 = 4.0 mM/min4.0\text{ mM/min}, Trial 2 = 4.6 mM/min4.6\text{ mM/min}, Trial 3 = 4.6 mM/min4.6\text{ mM/min}
The question specifically asks for the average at 35∘C35^\circ\text{C}.
2
Calculate the sum of the trial values
4.0+4.6+4.6=13.2 mM/min4.0 + 4.6 + 4.6 = 13.2\text{ mM/min}
Calculating an average requires finding the total sum of all trial measurements first.
3
Divide the sum by the total number of trials
13.23=4.4 mM/min\frac{13.2}{3} = 4.4\text{ mM/min}
Dividing the sum by 33 calculates the arithmetic mean across the 3 trials.

Key Concept

Calculating the average (mean) of data values from a table
Estimated Time:1m 0s
Question 5Question →

A team of plant physiologists investigated the transpiration rate of four plant species (Species W, X, Y, and Z) under two distinct light conditions—Low Intensity (150 μmol/m2/s150\text{ }\mu\text{mol/m}^2\text{/s}) and High Intensity (800 μmol/m2/s800\text{ }\mu\text{mol/m}^2\text{/s})—at a controlled temperature of 25∘C25^\circ\text{C}. The measured rates are shown in the table below:

Plant SpeciesTranspiration Rate at Low Intensity (mg H2O/cm2/hr\text{mg H}_2\text{O/cm}^2\text{/hr})Transpiration Rate at High Intensity (mg H2O/cm2/hr\text{mg H}_2\text{O/cm}^2\text{/hr})
Species W1.21.24.84.8
Species X2.52.56.56.5
Species Y0.80.83.23.2
Species Z1.51.55.55.5

Based on the table, what was the average increase in transpiration rate (in mg H2O/cm2/hr\text{mg H}_2\text{O/cm}^2\text{/hr}) across all four plant species when light condition was increased from Low Intensity to High Intensity?

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Answer: 3.5 mg H2O/cm2/hr3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

Answer

The average increase in transpiration rate across the four species is 3.5 mg H2O/cm2/hr3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}.
To find the average increase, first subtract the Low Intensity transpiration rate from the High Intensity transpiration rate for each of the four species: Species W (4.8−1.2=3.64.8 - 1.2 = 3.6), Species X (6.5−2.5=4.06.5 - 2.5 = 4.0), Species Y (3.2−0.8=2.43.2 - 0.8 = 2.4), and Species Z (5.5−1.5=4.05.5 - 1.5 = 4.0). Adding these four increases yields a total of 14.0 mg H2O/cm2/hr14.0\text{ mg H}_2\text{O/cm}^2\text{/hr}. Dividing this sum by 44 species gives an average increase of 3.5 mg H2O/cm2/hr3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}.

Step-by-Step Solution

1
Calculate the difference between High Intensity and Low Intensity transpiration rates for each species.
Species W: 4.8−1.2=3.6 mg H2O/cm2/hr4.8 - 1.2 = 3.6\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species X: 6.5−2.5=4.0 mg H2O/cm2/hr6.5 - 2.5 = 4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species Y: 3.2−0.8=2.4 mg H2O/cm2/hr3.2 - 0.8 = 2.4\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species Z: 5.5−1.5=4.0 mg H2O/cm2/hr5.5 - 1.5 = 4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}.
Determines the specific rate increase for each trial.
2
Sum the calculated rate increases across all four species.
3.6+4.0+2.4+4.0=14.0 mg H2O/cm2/hr3.6 + 4.0 + 2.4 + 4.0 = 14.0\text{ mg H}_2\text{O/cm}^2\text{/hr}.
Aggregates the individual increases to calculate the total change.
3
Divide the total sum of increases by the number of species (44) to determine the mean increase.
14.0/4=3.5 mg H2O/cm2/hr14.0 / 4 = 3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}.
Yields the average (mean) change in transpiration rate per species.

Key Concept

Calculating mean changes and rate differences from tabular data
Estimated Time:1m 0s
Question 6Question →

Astrophysicists measured the transit depth (percentage of starlight blocked) for four exoplanets orbiting a host star at two observation wavelengths (0.5 μm0.5\ \mu\text{m} and 1.5 μm1.5\ \mu\text{m}), as recorded in the table below:

ExoplanetTransit Depth at 0.5 μm0.5\ \mu\text{m} (%)Transit Depth at 1.5 μm1.5\ \mu\text{m} (%)
Exoplanet 11.21.8
Exoplanet 22.52.1
Exoplanet 30.81.6
Exoplanet 43.03.6

True or False: The mean transit depth across all four exoplanets measured at 1.5 μm1.5\ \mu\text{m} is 0.4%0.4\% greater than the mean transit depth measured at 0.5 μm0.5\ \mu\text{m}.

Show answer & explanation

Answer: True

Answer

The statement is True.
The mean transit depth at 1.5 μm is 2.275%, and the mean transit depth at 0.5 μm is 1.875%. Subtracting 1.875% from 2.275% gives exactly 0.4%, making the true/false statement correct.

Step-by-Step Solution

1
Calculate the mean transit depth at 0.5 μm.
Sum = 1.2 + 2.5 + 0.8 + 3.0 = 7.5%. Mean = 7.5 / 4 = 1.875%.
Determines the baseline average value for the first wavelength column.
2
Calculate the mean transit depth at 1.5 μm.
Sum = 1.8 + 2.1 + 1.6 + 3.6 = 9.1%. Mean = 9.1 / 4 = 2.275%.
Determines the target average value for the second wavelength column.
3
Calculate the difference between the two averages.
2.275% - 1.875% = 0.4%.
Compares the calculated difference against the claimed value of 0.4%.

Key Concept

Calculating and comparing average values from structured tabular data
Question 7Question →

A team of marine biologists studied the acoustic activity of the snapping shrimp (*Alpheus heterochaelis*) under controlled laboratory conditions. They measured the average snap rate (in snaps per minute) across three water temperatures and two salinity levels (30 PSU30\text{ PSU} and 36 PSU36\text{ PSU}).

Table 1
Water Temperature (∘C^\circ\text{C})Snap Rate at 30 PSU30\text{ PSU} (snaps/min)Snap Rate at 36 PSU36\text{ PSU} (snaps/min)
181845455050
222265658585
2626109109130130

Based on Table 1, what was the average rate of increase in snap rate (in snaps/min per 1∘C1^\circ\text{C}) for shrimp kept at a salinity of 36 PSU36\text{ PSU} as the water temperature increased from 18∘C18^\circ\text{C} to 26∘C26^\circ\text{C}?

Show answer & explanation

Answer: 10.0 snaps/min per ∘C10.0\text{ snaps/min per }^\circ\text{C}

Answer

The average rate of increase in snap rate for shrimp at 36 PSU36\text{ PSU} from 18∘C18^\circ\text{C} to 26∘C26^\circ\text{C} is 10.0 snaps/min per ∘C10.0\text{ snaps/min per }^\circ\text{C}.
To find the average rate of increase per 1∘C1^\circ\text{C}, determine the difference in snap rate at 36 PSU36\text{ PSU} between 18∘C18^\circ\text{C} and 26∘C26^\circ\text{C}, which is 130−50=80 snaps/min130 - 50 = 80\text{ snaps/min}. Then divide by the temperature span of 26−18=8∘C26 - 18 = 8^\circ\text{C}, yielding 80/8=10.0 snaps/min per ∘C80 / 8 = 10.0\text{ snaps/min per }^\circ\text{C}.

Step-by-Step Solution

1
Locate the snap rates for 36 PSU36\text{ PSU} at 18∘C18^\circ\text{C} and 26∘C26^\circ\text{C} in Table 1.
At 18∘C18^\circ\text{C}, snap rate = 50 snaps/min50\text{ snaps/min}; at 26∘C26^\circ\text{C}, snap rate = 130 snaps/min130\text{ snaps/min}.
The question asks specifically about the 36 PSU36\text{ PSU} salinity condition.
2
Calculate the total change in snap rate and total change in temperature.
Change in snap rate = 130−50=80 snaps/min130 - 50 = 80\text{ snaps/min}; Change in temperature = 26−18=8∘C26 - 18 = 8^\circ\text{C}.
Rate of change requires dividing the change in the dependent variable by the change in the independent variable.
3
Divide the change in snap rate by the change in temperature.
80 snaps/min/8∘C=10.0 snaps/min per ∘C80\text{ snaps/min} / 8^\circ\text{C} = 10.0\text{ snaps/min per }^\circ\text{C}.
This yields the average change per 1∘C1^\circ\text{C} increase.

Key Concept

Calculating rate of change from tabular data
Estimated Time:1m 15s
Question 8Question →

A group of biochemical researchers measured the enzymatic breakdown rate of cellulose (in mg/L⋅min\text{mg/L}\cdot\text{min}) by three bacterial strains (Strain X, Strain Y, and Strain Z) across three different pH environments (5.05.0, 6.06.0, and 7.07.0) at a constant temperature of 37∘C37^\circ\text{C}. The results are recorded in the table below:

StrainBreakdown Rate at pH 5.05.0Breakdown Rate at pH 6.06.0Breakdown Rate at pH 7.07.0
Strain X121218182424
Strain Y151525253535
Strain Z8814142020

True or False: The average enzymatic breakdown rate across all three pH environments for Strain Y is 5 mg/L⋅min5\text{ mg/L}\cdot\text{min} greater than the average breakdown rate for Strain X.

Show answer & explanation

Answer: False

Answer

The statement is False because the average breakdown rate for Strain Y (25 mg/L⋅min25\text{ mg/L}\cdot\text{min}) exceeds that of Strain X (18 mg/L⋅min18\text{ mg/L}\cdot\text{min}) by 7 mg/L⋅min7\text{ mg/L}\cdot\text{min}, not 5 mg/L⋅min5\text{ mg/L}\cdot\text{min}.
The correct response is False because computing the mean breakdown rates yields 18 mg/L⋅min18\text{ mg/L}\cdot\text{min} for Strain X and 25 mg/L⋅min25\text{ mg/L}\cdot\text{min} for Strain Y. Subtracting 1818 from 2525 gives a difference of 7 mg/L⋅min7\text{ mg/L}\cdot\text{min}, rendering the statement claiming a difference of 5 mg/L⋅min5\text{ mg/L}\cdot\text{min} mathematically incorrect.

Step-by-Step Solution

1
Calculate the average breakdown rate for Strain X across all three pH levels.
Average for Strain X = 12+18+243=543=18 mg/L⋅min\frac{12 + 18 + 24}{3} = \frac{54}{3} = 18\text{ mg/L}\cdot\text{min}.
To determine the overall mean performance of Strain X across the tested environments.
2
Calculate the average breakdown rate for Strain Y across all three pH levels.
Average for Strain Y = 15+25+353=753=25 mg/L⋅min\frac{15 + 25 + 35}{3} = \frac{75}{3} = 25\text{ mg/L}\cdot\text{min}.
To determine the overall mean performance of Strain Y across the tested environments.
3
Subtract the average rate of Strain X from the average rate of Strain Y.
Difference = 25−18=7 mg/L⋅min25 - 18 = 7\text{ mg/L}\cdot\text{min}.
To test the claim that the average for Strain Y is 5 mg/L⋅min5\text{ mg/L}\cdot\text{min} greater than Strain X.

Key Concept

Calculating mean values from tabular experimental data and finding differences between summary statistics.
Question 9Question →

Bioacousticians measured the echolocation click rate (in clicks per minute, clicks/min\text{clicks/min}) of a harbor porpoise at four different water depths across three 15-minute observation trials. The results are summarized in the table below:

Water Depth (m)Trial 1 (clicks/min)Trial 2 (clicks/min)Trial 3 (clicks/min)
10120115125
20140148138
30185170185
40210205215

Based on the table, what is the average echolocation click rate, in clicks/min\text{clicks/min}, across all three trials at a depth of 30 m?

Show answer & explanation

Answer: 180

Answer

The average echolocation click rate at a depth of 30 m across all three trials is 180 clicks/min.
To calculate the average echolocation click rate at 30 m depth, locate the row for 30 m, add the click rates from the three trials (185+170+185=540185 + 170 + 185 = 540), and divide by the number of trials (540÷3=180 clicks/min540 \div 3 = 180\text{ clicks/min}).

Step-by-Step Solution

1
Locate the depth row for 30 m in the provided data table.
Retrieved values for Trial 1 (185), Trial 2 (170), and Trial 3 (185).
The question specifically requests calculations for the 30 m depth.
2
Sum the recorded click rates across the three trials.
185+170+185=540 clicks/min185 + 170 + 185 = 540\text{ clicks/min}.
Obtaining the sum is required before dividing by the total count of trials.
3
Divide the calculated sum by 3.
5403=180 clicks/min\frac{540}{3} = 180\text{ clicks/min}.
Calculating the arithmetic mean yields the average click rate.

Key Concept

Calculating the arithmetic mean from tabular data
Estimated Time:1m 0s
Question 10Question →

A team of oceanographers monitored sea surface temperatures (SST, in ∘C{}^\circ\text{C}) at three distinct coral reef monitoring stations (S1S_1, S2S_2, and S3S_3) during a 4-month summer observation period.

Table 1:
StationJune SST (∘C{}^\circ\text{C})July SST (∘C{}^\circ\text{C})August SST (∘C{}^\circ\text{C})September SST (∘C{}^\circ\text{C})
S1S_127.528.529.528.5
S2S_228.029.030.031.0
S3S_326.527.028.528.0

Based on Table 1, what is the difference between the average 4-month SST at Station S2S_2 and the average 4-month SST at Station S3S_3?

Show answer & explanation

Answer: 2.0∘C2.0{}^\circ\text{C}

Answer

2.0∘C2.0{}^\circ\text{C}
To find the difference between the 4-month average SSTs at Station S₂ and Station S₃, first sum and average the monthly temperatures for each station. For Station S₂, the sum is 28.0 + 29.0 + 30.0 + 31.0 = 118.0 °C, and the average is 118.0 / 4 = 29.5 °C. For Station S₃, the sum is 26.5 + 27.0 + 28.5 + 28.0 = 110.0 °C, and the average is 110.0 / 4 = 27.5 °C. The difference between these two averages is 29.5 °C - 27.5 °C = 2.0 °C.

Step-by-Step Solution

1
Calculate the 4-month average SST for Station S2S_2.
Sum = 28.0+29.0+30.0+31.0=118.0∘C28.0 + 29.0 + 30.0 + 31.0 = 118.0{}^\circ\text{C}. Average = 118.0/4=29.5∘C118.0 / 4 = 29.5{}^\circ\text{C}.
Finding the mean value requires summing all four monthly measurements for Station S2S_2 and dividing by 4.
2
Calculate the 4-month average SST for Station S3S_3.
Sum = 26.5+27.0+28.5+28.0=110.0∘C26.5 + 27.0 + 28.5 + 28.0 = 110.0{}^\circ\text{C}. Average = 110.0/4=27.5∘C110.0 / 4 = 27.5{}^\circ\text{C}.
Finding the mean value requires summing all four monthly measurements for Station S3S_3 and dividing by 4.
3
Subtract the average SST of Station S3S_3 from the average SST of Station S2S_2.
29.5∘C−27.5∘C=2.0∘C29.5{}^\circ\text{C} - 27.5{}^\circ\text{C} = 2.0{}^\circ\text{C}.
Determining the difference between the two calculated averages.

Key Concept

Calculating average values across multiple data points in a table and finding their numerical difference.
Question 11Question →

Plant physiologists measured the photosynthetic rate (in μmol CO2/m2⋅s\mu\text{mol CO}_2/\text{m}^2\cdot\text{s}) of a C3 plant species exposed to three ambient temperature conditions (20∘C20^\circ\text{C}, 30∘C30^\circ\text{C}, and 40∘C40^\circ\text{C}) under constant saturating light intensity. The measured photosynthetic rates were 15.0 μmol CO2/m2⋅s15.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s} at 20∘C20^\circ\text{C}, 24.0 μmol CO2/m2⋅s24.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s} at 30∘C30^\circ\text{C}, and 9.0 μmol CO2/m2⋅s9.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s} at 40∘C40^\circ\text{C}.

Based on these data, is the following statement True or False?
"The mean photosynthetic rate of the plant across the three tested temperatures is 16.0 μmol CO2/m2⋅s16.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s}."

Show answer & explanation

Answer: True

Answer

True. The mean photosynthetic rate across all three tested temperatures is 16.0 μmol CO2/m2⋅s16.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s}.
The calculated average of the three photosynthetic rates (15.015.0, 24.024.0, and 9.09.0) is 15.0+24.0+9.03=48.03=16.0 μmol CO2/m2⋅s\frac{15.0 + 24.0 + 9.0}{3} = \frac{48.0}{3} = 16.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s}, which matches the claim directly.

Step-by-Step Solution

1
Extract the photosynthetic rates for all three temperature conditions.
Rate at 20∘C=15.020^\circ\text{C} = 15.0, Rate at 30∘C=24.030^\circ\text{C} = 24.0, Rate at 40∘C=9.040^\circ\text{C} = 9.0.
Identifying all data points is required to calculate the overall average.
2
Sum the rates recorded across the three conditions.
15.0+24.0+9.0=48.0 μmol CO2/m2⋅s15.0 + 24.0 + 9.0 = 48.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s}.
Finding the total sum is the first step in calculating an arithmetic mean.
3
Divide the total sum by the total number of conditions (n=3n = 3).
48.03=16.0 μmol CO2/m2⋅s\frac{48.0}{3} = 16.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s}.
Dividing the sum by the count yields the correct mean value.

Key Concept

Calculating Arithmetic Mean from Data Points
Calculating with Data Practice Questions — ACT | Examkin