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Let's solve this problem step by step. 

### Given:
- Each offspring pea has **0.25 probability** of having a green pod.
- The **total sample size is 28** (group size = 28).
- We're asked to find the **mean (μ)** and **standard deviation (σ)** for the number of peas with green pods.

---

### Step 1: Calculate the Mean (μ)
This is a **binomial distribution** problem. For a binomial distribution:
\[
\mu = n \cdot p
\]
where:
- \(n = 28\) (number of trials)
- \(p = 0.25\) (probability of success)

Now:
\[
\mu = 28 \cdot 0.25 = 7
\]

So, the mean is \( \mu = 7 \) peas.

---

### Step 2: Calculate the Standard Deviation (σ)
The formula for the **standard deviation (σ)** of a binomial distribution is:
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]

Substitute:
- \(n = 28\)
- \(p = 0.25\)
- \(1 - p = 0.75\)

\[
\sigma = \sqrt{28 \cdot 0.25 \cdot 0.75} = \sqrt{5.25} \approx 2.29
\]

---

### Final Answers:
- **Mean (μ)** = 7 peas
- **Standard Deviation (σ)** ≈ 2.29 peas

---

Would you like further explanation on any part? 

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### 5 Related Questions:
1. What is the probability of exactly 10 peas having green pods?
2. How does the binomial distribution behave when the probability \(p\) changes?
3. What happens to the mean if the group size increases to 50?
4. How is a normal approximation used for binomial distributions?
5. How can confidence intervals be applied to these results?

---

### Tip:
When dealing with binomial problems, always check if \(np(1 - p) \geq 10\) to see if a normal approximation can be applied!

Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.25 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 28. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 28.

Learn how to calculate the mean and standard deviation for the number of peas with green pods in a binomial distribution. The problem involves hybridization experiments where each pea has a 0.25 probability of having green pods, and groups of 28 peas are randomly selected.

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