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 peas are randomly selected in groups of 18. 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 18.
the value of the mean is μ = 13.5 peas.
(type an integer or a decimal. do not round.)
the value of the standard deviation is σ = peas.
(round to one decimal place as needed.)

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem, with \(n = 18\) (number of trials) and \(p=0.75\) (probability of success).

Step2: Recall the formula for standard - deviation of binomial distribution

The formula for the standard deviation of a binomial distribution is \(\sigma=\sqrt{np(1 - p)}\).

Step3: Substitute the values

Substitute \(n = 18\), \(p = 0.75\), and \(1-p=0.25\) into the formula: \(\sigma=\sqrt{18\times0.75\times0.25}\).

Step4: Calculate the value

First, \(18\times0.75\times0.25 = 18\times\frac{3}{4}\times\frac{1}{4}=\frac{54}{16}= 3.375\). Then \(\sigma=\sqrt{3.375}\approx1.8\).

Answer:

1.8