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?

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!