Let's solve this step-by-step based on the information provided.

Given Data:

• Mean weight, μ = 1395 grams
• Standard deviation, σ = 200 grams

Since the weights follow a normal distribution, we can use the z-score formula to convert these values for each part:

$z = \frac{X - \mu}{\sigma}$

Let's go through each part:

Part (a): Proportion of broilers weighing between 1160 and 1250 grams.

1. Calculate the z-score for 1160 grams: $z_{1160} = \frac{1160 - 1395}{200} = \frac{-235}{200} = -1.175$

2. Calculate the z-score for 1250 grams: $z_{1250} = \frac{1250 - 1395}{200} = \frac{-145}{200} = -0.725$

3. Find the corresponding probabilities from the cumulative normal distribution table:

   • $P(Z \leq -1.175) \approx 0.1203$
   • $P(Z \leq -0.725) \approx 0.2340$

4. Calculate the proportion between these z-scores: $P(-1.175 < Z < -0.725) = P(Z \leq -0.725) - P(Z \leq -1.175)$ $= 0.2340 - 0.1203 = 0.1137$

Therefore, the proportion of broilers that weigh between 1160 and 1250 grams is approximately 0.1137.

Would you like detailed solutions for parts (b) and (c), or have any questions about this part?

Here are 5 follow-up questions to consider:

1. How can z-scores be used to interpret normal distributions?
2. What does it mean if a z-score is positive or negative?
3. How can you find proportions for values outside the mean range?
4. What are the applications of normal distribution in quality control?
5. How does changing the standard deviation affect the distribution spread?

Tip: Always check the cumulative distribution table for precise values for z-scores, as small differences can affect final probabilities.