From the data shown in the table, the weighted mean size (in micrometer, correct to two decimal places) of the sediment population is ............
\[ \begin{array}{|c|c|} \hline \text{Grain Size (micrometer)} & \text{Dry Sediment Weight (in gram)} \\ \hline 4 & 50 \\ 20 & 75 \\ 40 & 125 \\ 60 & 50 \\ \hline \end{array} \]
From the data shown in the table, the weighted mean size (in micrometer, correct to two decimal places) of the sediment population is ............
\[ \begin{array}{|c|c|} \hline \text{Grain Size (micrometer)} & \text{Dry Sediment Weight (in gram)} \\ \hline 4 & 50 \\ 20 & 75 \\ 40 & 125 \\ 60 & 50 \\ \hline \end{array} \]
Show Hint
Correct Answer: 32 - 33
Solution and Explanation
Step 1: Calculate the total weight.
The total weight of the sediment population is the sum of the dry sediment weights:
\[
\text{Total weight} = 50 + 75 + 125 + 50 = 300 \, \text{grams}
\]
Step 2: Calculate the weighted sum of the grain sizes.
The weighted sum is the sum of the product of each grain size and its corresponding weight:
\[
\text{Weighted sum} = (4 \times 50) + (20 \times 75) + (40 \times 125) + (60 \times 50)
\]
\[
\text{Weighted sum} = 200 + 1500 + 5000 + 3000 = 9700
\]
Step 3: Calculate the weighted mean size.
The weighted mean size \( \overline{x} \) is given by the formula:
\[
\overline{x} = \frac{\text{Weighted sum}}{\text{Total weight}} = \frac{9700}{300} = 32.33 \, \mu m
\]
Thus, the weighted mean size of the sediment population is 32.33 micrometers.
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