Data Representation and Interpretation

23 questions

Question 21Question →

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 22Question →

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
Question 23Question →

Biophysicists conducted an experiment measuring the action potential conduction velocity (vv, in m/s\text{m/s}) in unmyelinated giant nerve fibers as a function of fiber diameter (dd, in μm\mu\text{m}) at a constant temperature of 20∘C20^\circ\text{C}. The recorded data is presented in Table 1.

Fiber Diameter (dd, μm\mu\text{m})Conduction Velocity (vv, m/s\text{m/s})
1005.0
2008.0
40014.0
60018.0
90021.0

Based on Table 1, what is the average rate of change in conduction velocity, in m/s\text{m/s} per 100 μm100\,\mu\text{m} increase in fiber diameter, over the interval from d=200 μmd = 200\,\mu\text{m} to d=600 μmd = 600\,\mu\text{m}?

Show answer & explanation

Answer: 2.5

Answer

The average rate of change in conduction velocity over the specified interval is 2.5 m/s per 100 μm2.5\,\text{m/s per }100\,\mu\text{m} increase in fiber diameter.
To find the average rate of change in conduction velocity per 100 μm100\,\mu\text{m} increase in fiber diameter between d=200 μmd = 200\,\mu\text{m} and d=600 μmd = 600\,\mu\text{m}, subtract the initial velocity (8.0 m/s8.0\,\text{m/s}) from the final velocity (18.0 m/s18.0\,\text{m/s}) to get Δv=10.0 m/s\Delta v = 10.0\,\text{m/s}. Divide by the change in diameter Δd=600−200=400 μm\Delta d = 600 - 200 = 400\,\mu\text{m} to obtain 0.025 m/s per μm0.025\,\text{m/s per }\mu\text{m}. Multiplying by 100100 yields 2.5 m/s per 100 μm2.5\,\text{m/s per }100\,\mu\text{m}.

Step-by-Step Solution

1
Locate data points for d=200 μmd = 200\,\mu\text{m} and d=600 μmd = 600\,\mu\text{m} in Table 1.
At d=200 μmd = 200\,\mu\text{m}, v=8.0 m/sv = 8.0\,\text{m/s}. At d=600 μmd = 600\,\mu\text{m}, v=18.0 m/sv = 18.0\,\text{m/s}.
These data points define the boundaries of the interval specified in the question.
2
Calculate the overall changes in velocity (Δv\Delta v) and diameter (Δd\Delta d).
Δv=18.0−8.0=10.0 m/s\Delta v = 18.0 - 8.0 = 10.0\,\text{m/s} and Δd=600−200=400 μm\Delta d = 600 - 200 = 400\,\mu\text{m}.
Calculating the rate of change requires dividing the change in the dependent variable by the change in the independent variable.
3
Scale the rate of change to a 100 μm100\,\mu\text{m} diameter interval.
10.0 m/s400 μm×100 μm=2.5 m/s per 100 μm\frac{10.0\,\text{m/s}}{400\,\mu\text{m}} \times 100\,\mu\text{m} = 2.5\,\text{m/s per }100\,\mu\text{m}.
The question asks specifically for the rate per 100 μm100\,\mu\text{m} increase in diameter.

Key Concept

Calculating average rate of change and trend slopes from quantitative scientific data tables.
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