The mean incubation time of fertilized eggs is 

2121 

days. Suppose the incubation times are approximately normally distributed with a standard deviation of 

11 

day.

​(a) Determine the 

1313th 

percentile for incubation times.

​(b) Determine the incubation times that make up the middle 

9797​%.

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Part 1

​(a) The 

1313th 

percentile for incubation times is 

______

days.

​(Round to the nearest whole number as​ needed.) 

We are dealing with a **normally distributed variable** representing incubation times. Let’s solve both parts step by step.

### **Given:**
- Mean \( \mu = 2121 \, \text{days} \)
- Standard deviation \( \sigma = 11 \, \text{days} \)

We will use the **standard normal distribution** (Z-distribution) to solve these problems.

---

## Part (a): 13th Percentile (P13)
To find the **13th percentile**, we need the Z-score associated with the 13th percentile in a standard normal distribution.

1. **Find the Z-score:**
   Using a Z-table or inverse Z-scores for percentiles, the Z-score for **13th percentile (P13)** is approximately:
   \[
   Z \approx -1.13
   \]
   This means that the 13th percentile lies 1.13 standard deviations below the mean.

2. **Convert Z-score to incubation time:**
   Use the formula to convert from Z-score to the raw value \( X \):
   \[
   X = \mu + Z \cdot \sigma
   \]
   Substituting the known values:
   \[
   X = 2121 + (-1.13) \cdot 11
   \]
   \[
   X = 2121 - 12.43 \approx 2109 \, \text{days}
   \]

So, the **13th percentile is approximately 2109 days**.

---

## Part (b): Incubation times for the middle 97%
For the middle 97%, the goal is to determine the **cutoff values** that leave **1.5%** on each side of the distribution (since \(100\% - 97\% = 3\%\), divided equally into tails).

1. **Find Z-scores:**
   Using a Z-table:
   - The **lower Z-score** for 1.5th percentile is approximately \( Z = -2.17 \).
   - The **upper Z-score** for 98.5th percentile is approximately \( Z = 2.17 \).

2. **Convert Z-scores to incubation times:**
   - **Lower bound:**
     \[
     X = 2121 + (-2.17) \cdot 11 = 2121 - 23.87 \approx 2097 \, \text{days}
     \]
   - **Upper bound:**
     \[
     X = 2121 + (2.17) \cdot 11 = 2121 + 23.87 \approx 2145 \, \text{days}
     \]

So, the **middle 97%** of incubation times are approximately **2097 to 2145 days**.

---

### **Final Answers:**
(a) The 13th percentile for incubation times is **2109 days**.  
(b) The incubation times that make up the middle 97% are **2097 to 2145 days**.

---

Do you want to explore the calculations further or have questions?  
Here are five related questions to expand your understanding:

1. How can we interpret a Z-score in practical terms?
2. What percentile corresponds to a Z-score of 0?  
3. How would the 13th percentile change if the mean incubation time were higher?  
4. What is the probability that an egg will hatch in fewer than 2100 days?  
5. How do changes in standard deviation affect the spread of a normal distribution?

**Tip:** Always sketch a bell curve when working with normal distributions—it helps visualize percentiles and probabilities effectively!

The mean incubation time of fertilized eggs is 2121 days. Suppose the incubation times are approximately normally distributed with a standard deviation of 11 days. (a) Determine the 13th percentile for incubation times. (b) Determine the incubation times that make up the middle 97%.

This problem explores how to calculate the 13th percentile and the middle 97% of incubation times using the normal distribution with a mean of 2121 days and a standard deviation of 11 days. Learn how to apply Z-scores and percentiles in a step-by-step solution. Suitable for Grades 10-12.

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