In this problem, we are given that the probability of a pea having green pods is $p = 0.75$, and we are considering groups of 12 peas. This situation follows a binomial distribution, where the number of trials $n = 12$ and the probability of success $p = 0.75$.

Part (a): Calculating the Mean and Standard Deviation

1. Mean ($\mu$) of a binomial distribution:

   The formula for the mean of a binomial distribution is: $\mu = n \times p$ Substituting the values: $\mu = 12 \times 0.75 = 9$ So, the mean is 9 peas.

2. Standard deviation ($\sigma$) of a binomial distribution:

   The formula for the standard deviation is: $\sigma = \sqrt{n \times p \times (1-p)}$ Substituting the values: $\sigma = \sqrt{12 \times 0.75 \times (1 - 0.75)} = \sqrt{12 \times 0.75 \times 0.25}$ $\sigma = \sqrt{12 \times 0.1875} = \sqrt{2.25} = 1.5$

Final Answer:

• The mean ($\mu$) is: $9$ peas.
• The standard deviation ($\sigma$) is: $1.5$ peas.

Would you like more details or have any other questions?

Related Questions:

1. What is the probability of exactly 8 peas having green pods in this scenario?
2. How do you calculate the variance of a binomial distribution?
3. What is the probability of having more than 10 peas with green pods in this group of 12?
4. How would these values change if the probability of a green pod was different, say 0.6?
5. What assumptions are made about the independence of the peas in this experiment?

Tip:

Always remember that in a binomial distribution, the mean tells you the expected number of successes, while the standard deviation indicates how much variation there is from the mean.