The problem you're working on involves analyzing pea offspring in a hybridization experiment, where the probability that a pea has green pods is 0.75, and you're asked to complete parts (a) and (b) based on the data provided.

Part (a)

1. Mean (µ):
   The mean number of peas with green pods in a group of 28 is given as 21. The mean is calculated using: $\mu = n \times p$ Where:

   • $n = 28$ (the number of peas in the group),
   • $p = 0.75$ (the probability of a pea having green pods).

   Therefore: $\mu = 28 \times 0.75 = 21$

2. Standard Deviation (σ):
   The standard deviation is provided as $\sigma = 2.3$ peas. It's calculated using the formula: $\sigma = \sqrt{n \times p \times (1 - p)}$ Substituting the values: $\sigma = \sqrt{28 \times 0.75 \times (1 - 0.75)} = \sqrt{28 \times 0.75 \times 0.25} \approx 2.3$

Part (b)

The "range rule of thumb" suggests that values more than 2 standard deviations away from the mean are considered significantly low or high.

To find the significantly low value, we calculate: $\text{Low value} = \mu - 2\sigma = 21 - 2 \times 2.3 = 21 - 4.6 = 16.4$ So, peas fewer than or equal to 16.4 are considered significantly low. Rounded to one decimal place, this value is 16.4.

Let me know if you'd like more details or have any further questions!

Here are 5 follow-up questions that build on this:

1. How would you calculate the significantly high value based on the range rule of thumb?
2. What would happen if the probability $p$ was 0.5 instead of 0.75? How would the mean and standard deviation change?
3. Can the number of peas in the group be changed to get a different standard deviation? What if $n = 50$?
4. How does the standard deviation influence the range of significantly low or high values?
5. What is the probability of having a group of peas with a number of green pods significantly higher than the mean?

Tip: The range rule of thumb can be a useful approximation to quickly identify outliers without needing to conduct a full probability analysis.