Questions: Determine the probability of a six-year-old rainbow trout being less than 385 millimeters long: Fish story: According to a report by the U.S. Fish and Wildife Service, the mean length of six-year-old rainbow trout in the Arolik River in Alaska is 484 millimeters with a standard deviation of 44 millimeters. Assume these lengths are normally distributed. It is or is not unusual because the probability of a six-year-old rainbow trout less than 385 millimeters long is

Solution Steps

To determine the probability of a six-year-old rainbow trout being less than 385 millimeters long, we first calculate the Z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X = 385 \) mm (the length we are interested in),
• \( \mu = 484 \) mm (the mean length),
• \( \sigma = 44 \) mm (the standard deviation).

Substituting the values, we have:

\[ z = \frac{385 - 484}{44} = \frac{-99}{44} \approx -2.25 \]

Next, we calculate the probability \( P \) that a six-year-old rainbow trout is less than 385 mm long. This is given by:

\[ P = \Phi(Z_{end}) - \Phi(Z_{start}) \]

In our case, \( Z_{end} = -2.25 \) and \( Z_{start} = -\infty \). Thus, we find:

\[ P = \Phi(-2.25) - \Phi(-\infty) \approx 0.0122 \]

To determine if this probability is unusual, we compare it to a common threshold of 0.05. Since:

\[ P = 0.0122 < 0.05 \]

we conclude that it is unusual for a six-year-old rainbow trout to be less than 385 mm long.

Final Answer