Question

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 26. 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 26.
the value of the mean is \\( \mu= \\) peas.
(type an integer or a decimal. do not round.)

Explanation:

Step1: Identify the distribution type

This is a binomial distribution problem. For a binomial distribution, the mean formula is \(\mu = np\), and the standard deviation formula is \(\sigma=\sqrt{np(1 - p)}\), where \(n\) is the number of trials and \(p\) is the probability of success in a single trial. Here, \(n = 26\) (group size) and \(p=0.75\) (probability of a pea having green pods).

Step2: Calculate the mean

Using the formula \(\mu=np\), substitute \(n = 26\) and \(p = 0.75\).
\(\mu=26\times0.75=19.5\)

Step3: Calculate the standard deviation

First, find \(1 - p=1 - 0.75 = 0.25\). Then use the formula \(\sigma=\sqrt{np(1 - p)}\).
\(\sigma=\sqrt{26\times0.75\times0.25}=\sqrt{4.875}\approx2.21\)

Answer:

The mean \(\mu = 19.5\) peas. The standard deviation \(\sigma\approx2.21\) peas.