Question
According to a report by the U.S. Fish and Wildlife Service, the mean length of six-year-old rainbow trout in the Arolik River in Alaska is 482 millimeters with a standard deviation of 42 millimeters. Assume these lengths are normally distributed.
(a) What proportion of six-year-old rainbow trout are less than 450 millimeters long?
(b) What proportion of six-year-old rainbow trout are between 410 and 510 millimeters long?
(c) Is it unusual for a six-year-old rainbow trout to be less than 413 millimeters long?
(a) What proportion of six-year-old rainbow trout are less than 450 millimeters long?
(b) What proportion of six-year-old rainbow trout are between 410 and 510 millimeters long?
(c) Is it unusual for a six-year-old rainbow trout to be less than 413 millimeters long?
Answers
GPT-4o mini
To solve these problems, we can use the properties of the normal distribution, specifically the standard normal distribution (Z-distribution). We will convert the lengths using the z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \(X\) is the value for which we want to find the z-score,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.
Given:
- \(\mu = 482\) mm
- \(\sigma = 42\) mm
### Part (a)
We want to find the proportion of six-year-old rainbow trout that are less than 450 mm long:
1. Calculate the z-score for \(X = 450\):
\[
Z = \frac{450 - 482}{42} = \frac{-32}{42} \approx -0.7619
\]
2. Look up this z-score in the standard normal distribution table (or use a calculator). The cumulative probability for \(Z = -0.76\) is approximately \(0.2236\).
Thus, the proportion of six-year-old rainbow trout less than 450 mm long is approximately \(0.2236\) or **22.36%**.
### Part (b)
We want to find the proportion of six-year-old rainbow trout that are between 410 and 510 mm long:
1. Calculate the z-score for \(X = 410\):
\[
Z_{410} = \frac{410 - 482}{42} = \frac{-72}{42} \approx -1.7143
\]
2. Calculate the z-score for \(X = 510\):
\[
Z_{510} = \frac{510 - 482}{42} = \frac{28}{42} \approx 0.6667
\]
3. Look up these z-scores in the standard normal distribution table (or use a calculator):
- For \(Z_{410} \approx -1.71\), the cumulative probability is approximately \(0.0436\).
- For \(Z_{510} \approx 0.67\), the cumulative probability is approximately \(0.7486\).
4. Find the proportion between these two values:
\[
P(410 < X < 510) = P(Z < 0.67) - P(Z < -1.71) \approx 0.7486 - 0.0436 = 0.7050
\]
Thus, the proportion of six-year-old rainbow trout between 410 and 510 mm long is approximately \(0.7050\) or **70.50%**.
### Part (c)
We want to determine whether it is unusual for a six-year-old rainbow trout to be less than 413 mm long:
1. Calculate the z-score for \(X = 413\):
\[
Z = \frac{413 - 482}{42} = \frac{-69}{42} \approx -1.6429
\]
2. Look up this z-score (or use a calculator). The cumulative probability for \(Z \approx -1.64\) is approximately \(0.0505\).
A value is typically considered unusual if it is more than two standard deviations away from the mean (corresponding to approximately 95% of the distribution). Since \(P(X < 413) \approx 0.0505\), this means that about **5.05%** of the trout are less than 413 mm long, which is below 5% (more than two standard deviations from the mean).
Thus, it **is unusual** for a six-year-old rainbow trout to be less than 413 mm long.
\[
Z = \frac{X - \mu}{\sigma}
\]
where:
- \(X\) is the value for which we want to find the z-score,
- \(\mu\) is the mean,
- \(\sigma\) is the standard deviation.
Given:
- \(\mu = 482\) mm
- \(\sigma = 42\) mm
### Part (a)
We want to find the proportion of six-year-old rainbow trout that are less than 450 mm long:
1. Calculate the z-score for \(X = 450\):
\[
Z = \frac{450 - 482}{42} = \frac{-32}{42} \approx -0.7619
\]
2. Look up this z-score in the standard normal distribution table (or use a calculator). The cumulative probability for \(Z = -0.76\) is approximately \(0.2236\).
Thus, the proportion of six-year-old rainbow trout less than 450 mm long is approximately \(0.2236\) or **22.36%**.
### Part (b)
We want to find the proportion of six-year-old rainbow trout that are between 410 and 510 mm long:
1. Calculate the z-score for \(X = 410\):
\[
Z_{410} = \frac{410 - 482}{42} = \frac{-72}{42} \approx -1.7143
\]
2. Calculate the z-score for \(X = 510\):
\[
Z_{510} = \frac{510 - 482}{42} = \frac{28}{42} \approx 0.6667
\]
3. Look up these z-scores in the standard normal distribution table (or use a calculator):
- For \(Z_{410} \approx -1.71\), the cumulative probability is approximately \(0.0436\).
- For \(Z_{510} \approx 0.67\), the cumulative probability is approximately \(0.7486\).
4. Find the proportion between these two values:
\[
P(410 < X < 510) = P(Z < 0.67) - P(Z < -1.71) \approx 0.7486 - 0.0436 = 0.7050
\]
Thus, the proportion of six-year-old rainbow trout between 410 and 510 mm long is approximately \(0.7050\) or **70.50%**.
### Part (c)
We want to determine whether it is unusual for a six-year-old rainbow trout to be less than 413 mm long:
1. Calculate the z-score for \(X = 413\):
\[
Z = \frac{413 - 482}{42} = \frac{-69}{42} \approx -1.6429
\]
2. Look up this z-score (or use a calculator). The cumulative probability for \(Z \approx -1.64\) is approximately \(0.0505\).
A value is typically considered unusual if it is more than two standard deviations away from the mean (corresponding to approximately 95% of the distribution). Since \(P(X < 413) \approx 0.0505\), this means that about **5.05%** of the trout are less than 413 mm long, which is below 5% (more than two standard deviations from the mean).
Thus, it **is unusual** for a six-year-old rainbow trout to be less than 413 mm long.