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: 2.0C2.0{}^\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 22Question

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
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 20C20^\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 100μm100\,\mu\text{m} increase in fiber diameter, over the interval from d=200μmd = 200\,\mu\text{m} to d=600μmd = 600\,\mu\text{m}?

Show answer & explanation

Answer: 2.5

Answer

The average rate of change in conduction velocity over the specified interval is 2.5m/s per 100μm2.5\,\text{m/s per }100\,\mu\text{m} increase in fiber diameter.
To find the average rate of change in conduction velocity per 100μm100\,\mu\text{m} increase in fiber diameter between d=200μmd = 200\,\mu\text{m} and d=600μmd = 600\,\mu\text{m}, subtract the initial velocity (8.0m/s8.0\,\text{m/s}) from the final velocity (18.0m/s18.0\,\text{m/s}) to get Δv=10.0m/s\Delta v = 10.0\,\text{m/s}. Divide by the change in diameter Δd=600200=400μm\Delta d = 600 - 200 = 400\,\mu\text{m} to obtain 0.025m/s per μm0.025\,\text{m/s per }\mu\text{m}. Multiplying by 100100 yields 2.5m/s per 100μm2.5\,\text{m/s per }100\,\mu\text{m}.

Step-by-Step Solution

1
Locate data points for d=200μmd = 200\,\mu\text{m} and d=600μmd = 600\,\mu\text{m} in Table 1.
At d=200μmd = 200\,\mu\text{m}, v=8.0m/sv = 8.0\,\text{m/s}. At d=600μmd = 600\,\mu\text{m}, v=18.0m/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=18.08.0=10.0m/s\Delta v = 18.0 - 8.0 = 10.0\,\text{m/s} and Δd=600200=400μm\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 100μm100\,\mu\text{m} diameter interval.
10.0m/s400μm×100μm=2.5m/s per 100μm\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 100μm100\,\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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