To solve these problems, we will use the formula for calculating the z-score:
\[ z = \frac{(X - \mu)}{\sigma} \]
where:
- \( z \) is the z-score,
- \( X \) is the value in question,
- \( \mu \) is the mean of the distribution, and
- \( \sigma \) is the standard deviation.
A)
For Anna's one-year-old flounder:
- Mean (\( \mu \)) = 132 mm
- Standard Deviation (\( \sigma \)) = 20 mm
- Length caught (\( X \)) = 145 mm
Calculating the z-score:
\[ z = \frac{(145 - 132)}{20} = \frac{13}{20} = 0.65 \]
Answer A: The z-score for Anna's flounder is 0.65.
B)
For Luis's two-year-old flounder:
- Mean (\( \mu \)) = 161 mm
- Standard Deviation (\( \sigma \)) = 25 mm
- Length caught (\( X \)) = 195 mm
Calculating the z-score:
\[ z = \frac{(195 - 161)}{25} = \frac{34}{25} = 1.36 \]
Answer B: The z-score for Luis's flounder is 1.36.
C)
For Joe's one-year-old flounder with a z-score of 1.4: We need to find the length \( X \):
Given:
- Mean (\( \mu \)) = 132 mm
- Standard Deviation (\( \sigma \)) = 20 mm
- \( z = 1.4 \)
Using the z-score formula rearranged to solve for \( X \):
\[ X = z \cdot \sigma + \mu \]
Substituting in the values:
\[ X = (1.4 \cdot 20) + 132 = 28 + 132 = 160 \]
Answer C: Joe's flounder was 160.0 mm long.
D)
For Terry's two-year-old flounder with a z-score of -0.4: Again, we need to find the length \( X \):
Given:
- Mean (\( \mu \)) = 161 mm
- Standard Deviation (\( \sigma \)) = 25 mm
- \( z = -0.4 \)
Using the z-score formula rearranged to solve for \( X \):
\[ X = z \cdot \sigma + \mu \]
Substituting in the values:
\[ X = (-0.4 \cdot 25) + 161 = -10 + 161 = 151 \]
Answer D: Terry's flounder was 151.0 mm long.