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.

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.