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The problem deals with a binomial distribution based on a probability of a pea having green pods (0.75) and a sample size of 28 peas. Let's break it down:

### Part (a): Finding the Mean and Standard Deviation
- **Mean (μ)**:
  \[
  \mu = n \cdot p = 28 \cdot 0.75 = 21
  \]
  Therefore, the mean is **21 peas**.

- **Standard Deviation (σ)**:
  The formula for the standard deviation of a binomial distribution is:
  \[
  \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{28 \cdot 0.75 \cdot (1 - 0.75)} = \sqrt{28 \cdot 0.75 \cdot 0.25} = \sqrt{5.25} \approx 2.29
  \]
  Rounded to one decimal place, the standard deviation is **2.3 peas**.

### Part (b): Using the Range Rule of Thumb
The range rule of thumb states that values more than 2 standard deviations away from the mean are considered significantly low or high. We calculate the boundaries as follows:

- **Significantly Low**:
  \[
  \mu - 2\sigma = 21 - 2(2.3) = 21 - 4.6 = 16.4
  \]
  Therefore, values **16.4 peas or fewer** are considered significantly low.

- **Significantly High**:
  \[
  \mu + 2\sigma = 21 + 2(2.3) = 21 + 4.6 = 25.6
  \]
  Therefore, values **25.6 peas or greater** are considered significantly high.

### Part (c): Determining if 24 Peas is a Significantly High Result
We now check if 24 peas falls into the "significantly high" category:
- Since **24** peas lies between **16.4** and **25.6**, it is **not significantly high**.

Thus, the final answer for part (c) is:
- **The result is not significantly high because 24 peas with green pods is within the expected range of values.**

---

Would you like further clarification on any part of this problem? Here are five related questions to explore:

1. How is the probability of 0.75 determined in binomial experiments?
2. How would changing the group size (n) from 28 to 40 peas affect the mean and standard deviation?
3. What does the term "significantly low" mean in statistical contexts?
4. How would you calculate the probability of getting exactly 24 peas with green pods?
5. How could the standard deviation change if the probability were different, such as 0.5?

### Tip:
In a binomial distribution, the standard deviation gives an idea of how much variability there is around the mean, so the farther a result is from the mean (in terms of standard deviations), the more unusual it is!

Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 28. Complete parts (a) through (c) below.

Solve a binomial distribution problem with a probability of 0.75 for green pods in peas. Learn how to calculate the mean (21 peas), standard deviation (2.3 peas), and determine if a result is significantly high or low using the range rule of thumb.

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