The situation described in the problem involves a binomial distribution, where each pea has a probability of 0.75 of having green pods, and 20 peas are randomly selected (which corresponds to $n = 20$).

a. Finding the mean and standard deviation:

For a binomial distribution, the mean ($\mu$) and standard deviation ($\sigma$) are given by the following formulas:

• Mean: $\mu = n \cdot p$
• Standard Deviation: $\sigma = \sqrt{n \cdot p \cdot (1 - p)}$

Where:

• $n = 20$ (the number of trials or peas),
• $p = 0.75$ (the probability of a pea having green pods).

Now, let’s calculate the values.

Mean:

$\mu = 20 \cdot 0.75 = 15$

Standard deviation:

$\sigma = \sqrt{20 \cdot 0.75 \cdot (1 - 0.75)} = \sqrt{20 \cdot 0.75 \cdot 0.25} = \sqrt{3.75} \approx 1.936$

Summary:

• Mean: 15 peas
• Standard Deviation: Approximately 1.936 peas

Would you like a deeper explanation of binomial distributions? Here are some related questions for further exploration:

1. How does changing the probability $p$ affect the mean and standard deviation?
2. What would the distribution look like if we increased the group size to 50 peas?
3. What happens to the standard deviation if the probability was closer to 0.5?
4. Can we use the normal distribution to approximate this binomial distribution?
5. How does this relate to real-world hybridization experiments?

Tip: For large $n$, binomial distributions can often be approximated by a normal distribution, especially when $p$ is not too close to 0 or 1.