Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods. Assume that the offspring peas are randomly selected in groups of 36 . Complete parts (a) through (c) below.
a. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 36 .
The value of the mean is $\mu=\square$ peas.
(Type an integer or a decimal. Do not round.)

Step 1 ：Given that the probability of a pea having green pods is 0.75 and the peas are selected in groups of 36, we can use the properties of the binomial distribution to find the mean and standard deviation.

Step 2 ：The mean of a binomial distribution is given by the formula \(\mu = np\), where n is the number of trials and p is the probability of success. In this case, n = 36 (the number of peas in each group) and p = 0.75 (the probability that a pea has green pods).

Step 3 ：Substituting the given values into the formula, we get \(\mu = 36 \times 0.75 = 27\). So, the mean number of peas with green pods in groups of 36 is \(\boxed{27}\) peas.

Step 4 ：The standard deviation of a binomial distribution is given by the formula \(\sigma = \sqrt{np(1-p)}\), where n is the number of trials, p is the probability of success, and \(\sqrt{}\) is the square root function.

Step 5 ：Substituting the given values into the formula, we get \(\sigma = \sqrt{36 \times 0.75 \times (1-0.75)} = 2.598076211353316\).

Step 6 ：Rounding to two decimal places, the standard deviation is approximately \(\boxed{2.60}\) peas.