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The problem describes the weights of adult trout following a **Normal distribution**, and we're tasked with finding the **mean** (μ) and **standard deviation** (σ) that define this model.

Here’s what we know from the problem:

1. **22%** of the trout weigh **below 2 pounds**.
2. **6%** of the trout weigh **above 5 pounds**.

### Step 1: Use Z-scores
In a normal distribution, the percentage of data less than or greater than a given value corresponds to a **Z-score**. We can use a Z-table (or calculator) to find the Z-scores corresponding to these percentages.

- **22% below 2 pounds** means that the cumulative probability is 0.22. The Z-score for 0.22 is approximately **-0.77**.
  
- **6% above 5 pounds** means that the cumulative probability below 5 pounds is **0.94** (since 100% - 6% = 94%). The Z-score for 0.94 is approximately **1.55**.

### Step 2: Set up equations
The formula for a Z-score is:

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

We can now set up two equations based on the Z-scores and the weights:

1. For 2 pounds (Z = -0.77):
   \[
   -0.77 = \frac{2 - \mu}{\sigma}
   \]

2. For 5 pounds (Z = 1.55):
   \[
   1.55 = \frac{5 - \mu}{\sigma}
   \]

### Step 3: Solve the system of equations
We now solve this system of equations to find μ (mean) and σ (standard deviation).

#### First equation:
\[
\mu = 2 + 0.77\sigma \tag{1}
\]

#### Second equation:
\[
\mu = 5 - 1.55\sigma \tag{2}
\]

Set equation (1) equal to equation (2):

\[
2 + 0.77\sigma = 5 - 1.55\sigma
\]

Simplify and solve for σ:

\[
0.77\sigma + 1.55\sigma = 5 - 2
\]
\[
2.32\sigma = 3
\]
\[
\sigma = \frac{3}{2.32} \approx 1.29
\]

### Step 4: Find μ
Substitute σ = 1.29 into either equation (1) or (2). Using equation (1):

\[
\mu = 2 + 0.77(1.29) \approx 2 + 0.993 \approx 2.993
\]

So, the **mean** is approximately **2.99 pounds**, and the **standard deviation** is approximately **1.29 pounds**.

### Final Answer:
- **Mean (μ):** 2.99 pounds
- **Standard deviation (σ):** 1.29 pounds

---

Would you like more details or have any questions? Here are 5 follow-up questions:

1. How would the model change if the percentage of trout below 2 pounds increased?
2. What does the Z-score tell us about the distribution of the data?
3. How would we interpret the mean in this context for a biologist?
4. Could you calculate the probability that a randomly caught trout weighs exactly 3 pounds?
5. What happens to the model if the standard deviation increases?

**Tip:** Z-scores can be very helpful in quickly relating percentages to real values in normally distributed data!

Wildlife biologists believe that the weights of adult trout can be described by a Normal model. They collect data from fishermen, finding 22% of the trout caught were thrown back because they were below the 2-pound minimum, and only 6% weighed over 5 pounds. What mean and standard deviation should define the model?

Solve a problem involving trout weights modeled by a normal distribution. Learn how to calculate the mean and standard deviation given that 22% of trout weigh below 2 pounds and 6% weigh above 5 pounds. A practical example of applying Z-scores in statistics.

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