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 = 19.5 \\) peas.
(type an integer or a decimal. do not round.)
the value of the standard deviation is \\( \sigma = \square \\) peas.
(round to one decimal place as needed.)

Explanation:

Step1: Identify the distribution type

This is a binomial distribution problem. For a binomial distribution, the mean is given by \(\mu = np\) and the standard deviation is given by \(\sigma=\sqrt{np(1 - p)}\). Here, \(n = 26\) (number of trials, i.e., number of peas in a group) and \(p=0.75\) (probability of a pea having green pods).

Step2: Calculate the standard deviation

Substitute \(n = 26\) and \(p = 0.75\) into the formula for the standard deviation \(\sigma=\sqrt{np(1 - p)}\).
First, calculate \(np(1 - p)\):

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Then, find \(\sigma=\sqrt{4.875}\approx2.2\)

Answer:

\(2.2\)