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 25C25^\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/mLhr\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/mLhr\text{cells/mL}\cdot\text{hr})Trial 2 (cells/mLhr\text{cells/mL}\cdot\text{hr})Trial 3 (cells/mLhr\text{cells/mL}\cdot\text{hr})
25140155125
30280310250
35420460380
40210240180

True or False: The mean growth rate at 35C35^\circ\text{C} exceeds the mean growth rate at 30C30^\circ\text{C} by more than 130 cells/mLhr130\text{ cells/mL}\cdot\text{hr}.

Show answer & explanation

Answer: True

Answer

The statement is True because the mean growth rate at 35C35^\circ\text{C} (420 cells/mLhr420\text{ cells/mL}\cdot\text{hr}) exceeds the mean growth rate at 30C30^\circ\text{C} (280 cells/mLhr280\text{ cells/mL}\cdot\text{hr}) by 140 cells/mLhr140\text{ cells/mL}\cdot\text{hr}, which is greater than 130 cells/mLhr130\text{ cells/mL}\cdot\text{hr}.
The mean growth rate at 35C35^\circ\text{C} is 420 cells/mLhr420\text{ cells/mL}\cdot\text{hr} and at 30C30^\circ\text{C} is 280 cells/mLhr280\text{ cells/mL}\cdot\text{hr}. The difference is 140 cells/mLhr140\text{ cells/mL}\cdot\text{hr}, which is strictly greater than 130 cells/mLhr130\text{ cells/mL}\cdot\text{hr}.

Step-by-Step Solution

1
Calculate the average growth rate for 35C35^\circ\text{C} across all three trials.
Mean35=420+460+3803=12603=420 cells/mLhr\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 35C35^\circ\text{C} temperature condition requires finding the mean.
2
Calculate the average growth rate for 30C30^\circ\text{C} across all three trials.
Mean30=280+310+2503=8403=280 cells/mLhr\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 30C30^\circ\text{C} temperature condition requires finding the mean.
3
Subtract the mean rate at 30C30^\circ\text{C} from the mean rate at 35C35^\circ\text{C}.
420280=140 cells/mLhr420 - 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/mLhr140\text{ cells/mL}\cdot\text{hr}) to the value given in the statement (130 cells/mLhr130\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 5.53.0=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/day3.0 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 (25C25^\circ\text{C}, 35C35^\circ\text{C}, and 45C45^\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 35C35^\circ\text{C} treatment across the 3 trials?

Show answer & explanation

Answer: 4.4

Answer

The average rate of glucose consumption at 35C35^\circ\text{C} across the 3 trials is 4.4 mM/min4.4\text{ mM/min}.
To find the average glucose consumption rate at 35C35^\circ\text{C}, locate the row for 35C35^\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 35C35^\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 35C35^\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 25C25^\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?

Show answer & explanation

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 (4.81.2=3.64.8 - 1.2 = 3.6), Species X (6.52.5=4.06.5 - 2.5 = 4.0), Species Y (3.20.8=2.43.2 - 0.8 = 2.4), and Species Z (5.51.5=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: 4.81.2=3.6 mg H2O/cm2/hr4.8 - 1.2 = 3.6\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species X: 6.52.5=4.0 mg H2O/cm2/hr6.5 - 2.5 = 4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species Y: 3.20.8=2.4 mg H2O/cm2/hr3.2 - 0.8 = 2.4\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species Z: 5.51.5=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 1C1^\circ\text{C}) for shrimp kept at a salinity of 36 PSU36\text{ PSU} as the water temperature increased from 18C18^\circ\text{C} to 26C26^\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 18C18^\circ\text{C} to 26C26^\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 1C1^\circ\text{C}, determine the difference in snap rate at 36 PSU36\text{ PSU} between 18C18^\circ\text{C} and 26C26^\circ\text{C}, which is 13050=80 snaps/min130 - 50 = 80\text{ snaps/min}. Then divide by the temperature span of 2618=8C26 - 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 18C18^\circ\text{C} and 26C26^\circ\text{C} in Table 1.
At 18C18^\circ\text{C}, snap rate = 50 snaps/min50\text{ snaps/min}; at 26C26^\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 = 13050=80 snaps/min130 - 50 = 80\text{ snaps/min}; Change in temperature = 2618=8C26 - 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/8C=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 1C1^\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/Lmin\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 37C37^\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/Lmin5\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/Lmin25\text{ mg/L}\cdot\text{min}) exceeds that of Strain X (18 mg/Lmin18\text{ mg/L}\cdot\text{min}) by 7 mg/Lmin7\text{ mg/L}\cdot\text{min}, not 5 mg/Lmin5\text{ mg/L}\cdot\text{min}.
The correct response is False because computing the mean breakdown rates yields 18 mg/Lmin18\text{ mg/L}\cdot\text{min} for Strain X and 25 mg/Lmin25\text{ mg/L}\cdot\text{min} for Strain Y. Subtracting 1818 from 2525 gives a difference of 7 mg/Lmin7\text{ mg/L}\cdot\text{min}, rendering the statement claiming a difference of 5 mg/Lmin5\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/Lmin\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/Lmin\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 = 2518=7 mg/Lmin25 - 18 = 7\text{ mg/L}\cdot\text{min}.
To test the claim that the average for Strain Y is 5 mg/Lmin5\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: 2.0C2.0{}^\circ\text{C}

Answer

2.0C2.0{}^\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=118.0C28.0 + 29.0 + 30.0 + 31.0 = 118.0{}^\circ\text{C}. Average = 118.0/4=29.5C118.0 / 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=110.0C26.5 + 27.0 + 28.5 + 28.0 = 110.0{}^\circ\text{C}. Average = 110.0/4=27.5C110.0 / 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.
29.5C27.5C=2.0C29.5{}^\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/m2s\mu\text{mol CO}_2/\text{m}^2\cdot\text{s}) of a C3 plant species exposed to three ambient temperature conditions (20C20^\circ\text{C}, 30C30^\circ\text{C}, and 40C40^\circ\text{C}) under constant saturating light intensity. The measured photosynthetic rates were 15.0 μmol CO2/m2s15.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s} at 20C20^\circ\text{C}, 24.0 μmol CO2/m2s24.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s} at 30C30^\circ\text{C}, and 9.0 μmol CO2/m2s9.0\ \mu\text{mol CO}_2/\text{m}^2\cdot\text{s} at 40C40^\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/m2s16.0\ \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/m2s16.0\ \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/m2s\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 20C=15.020^\circ\text{C} = 15.0, Rate at 30C=24.030^\circ\text{C} = 24.0, Rate at 40C=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/m2s15.0 + 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/m2s\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