Question
The mean length of one-year-old spotted flounder, in millimeters, is 132 with standard deviation of 20, and the mean length of two-year-old spotted flounder is 161 with a standard deviation of 25 . The distribution of flounder lengths is approximately bell-shaped.
A) Anna caught a one-year-old flounder that was 145 millimeters in length. What is the -score for this length? Round the answers to at least two decimal places.
B) (b) Luis caught a two-year-old flounder that was 195 millimeters in length. What is the -score for this length? Round the answers to at least two decimal places.
C) Joe caught a one-year-old flounder whose length had a -score of 1.4. How long was this fish? Round the answer to at least one decimal place.
D) Terry caught a two-year-old flounder whose length had a -score of -0.4 . How long was this fish? Round the answer to at least one decimal place.
A) Anna caught a one-year-old flounder that was 145 millimeters in length. What is the -score for this length? Round the answers to at least two decimal places.
B) (b) Luis caught a two-year-old flounder that was 195 millimeters in length. What is the -score for this length? Round the answers to at least two decimal places.
C) Joe caught a one-year-old flounder whose length had a -score of 1.4. How long was this fish? Round the answer to at least one decimal place.
D) Terry caught a two-year-old flounder whose length had a -score of -0.4 . How long was this fish? Round the answer to at least one decimal place.
Answers
GPT-4o mini
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.
\[
z = \frac{(X - \mu)}{\sigma}
\]
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.