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!