Answer :
To determine the distribution pattern of oyster larvae across four different sites over three days, we begin by calculating the mean density and variance of the larvae density at each site. Distribution patterns can often be inferred from the variance, where high variance might suggest a clumped distribution.
Let's break down the calculations step-by-step:
### Step 1: Compute the Mean Density
We calculate the mean density for each site (W, X, Y, Z) across the three days.
Site W:
[tex]\[ \text{Mean Density (W)} = \frac{40 + 3 + 2}{3} = \frac{45}{3} = 15.0 \][/tex]
Site X:
[tex]\[ \text{Mean Density (X)} = \frac{22 + 1 + 6}{3} = \frac{29}{3} \approx 9.67 \][/tex]
Site Y:
[tex]\[ \text{Mean Density (Y)} = \frac{0 + 14 + 3}{3} = \frac{17}{3} \approx 5.67 \][/tex]
Site Z:
[tex]\[ \text{Mean Density (Z)} = \frac{7 + 26 + 1}{3} = \frac{34}{3} \approx 11.33 \][/tex]
### Step 2: Compute the Variance
Next, we calculate the variance for each site, which measures how much the densities at each site vary from their respective means.
Site W:
[tex]\[ \text{Variance (W)} = \frac{(40 - 15)^2 + (3 - 15)^2 + (2 - 15)^2}{3} = \frac{625 + 144 + 169}{3} = \frac{938}{3} \approx 312.67 \][/tex]
Site X:
[tex]\[ \text{Variance (X)} = \frac{(22 - 9.67)^2 + (1 - 9.67)^2 + (6 - 9.67)^2}{3} = \frac{153.51 + 75.11 + 13.44}{3} \approx 80.22 \][/tex]
Site Y:
[tex]\[ \text{Variance (Y)} = \frac{(0 - 5.67)^2 + (14 - 5.67)^2 + (3 - 5.67)^2}{3} = \frac{32.11 + 69.44 + 7.11}{3} \approx 36.22 \][/tex]
Site Z:
[tex]\[ \text{Variance (Z)} = \frac{(7 - 11.33)^2 + (26 - 11.33)^2 + (1 - 11.33)^2}{3} = \frac{18.78 + 214.78 + 106.78}{3} = 113.56 \][/tex]
### Step 3: Interpret the Variances
We analyze the variances to infer the distribution pattern:
- Site W: 312.67
- Site X: 80.22
- Site Y: 36.22
- Site Z: 113.56
A high variance often suggests a clumped distribution because there are larger discrepancies in the density readings.
### Conclusion
Given the variances at the different sites, particularly the highest variance being 312.67 at Site W, it can be inferred that the distribution pattern of the oyster larvae is clumped.
Thus, the answer is:
C. clumped
Let's break down the calculations step-by-step:
### Step 1: Compute the Mean Density
We calculate the mean density for each site (W, X, Y, Z) across the three days.
Site W:
[tex]\[ \text{Mean Density (W)} = \frac{40 + 3 + 2}{3} = \frac{45}{3} = 15.0 \][/tex]
Site X:
[tex]\[ \text{Mean Density (X)} = \frac{22 + 1 + 6}{3} = \frac{29}{3} \approx 9.67 \][/tex]
Site Y:
[tex]\[ \text{Mean Density (Y)} = \frac{0 + 14 + 3}{3} = \frac{17}{3} \approx 5.67 \][/tex]
Site Z:
[tex]\[ \text{Mean Density (Z)} = \frac{7 + 26 + 1}{3} = \frac{34}{3} \approx 11.33 \][/tex]
### Step 2: Compute the Variance
Next, we calculate the variance for each site, which measures how much the densities at each site vary from their respective means.
Site W:
[tex]\[ \text{Variance (W)} = \frac{(40 - 15)^2 + (3 - 15)^2 + (2 - 15)^2}{3} = \frac{625 + 144 + 169}{3} = \frac{938}{3} \approx 312.67 \][/tex]
Site X:
[tex]\[ \text{Variance (X)} = \frac{(22 - 9.67)^2 + (1 - 9.67)^2 + (6 - 9.67)^2}{3} = \frac{153.51 + 75.11 + 13.44}{3} \approx 80.22 \][/tex]
Site Y:
[tex]\[ \text{Variance (Y)} = \frac{(0 - 5.67)^2 + (14 - 5.67)^2 + (3 - 5.67)^2}{3} = \frac{32.11 + 69.44 + 7.11}{3} \approx 36.22 \][/tex]
Site Z:
[tex]\[ \text{Variance (Z)} = \frac{(7 - 11.33)^2 + (26 - 11.33)^2 + (1 - 11.33)^2}{3} = \frac{18.78 + 214.78 + 106.78}{3} = 113.56 \][/tex]
### Step 3: Interpret the Variances
We analyze the variances to infer the distribution pattern:
- Site W: 312.67
- Site X: 80.22
- Site Y: 36.22
- Site Z: 113.56
A high variance often suggests a clumped distribution because there are larger discrepancies in the density readings.
### Conclusion
Given the variances at the different sites, particularly the highest variance being 312.67 at Site W, it can be inferred that the distribution pattern of the oyster larvae is clumped.
Thus, the answer is:
C. clumped