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 30.
find the mean and the standard deviation for the numbers of peas with green pods in the groups of 30.
the value of the mean is μ = □ 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: Calculate the mean

For a binomial distribution, the mean formula is \(\mu = np\). Here, \(n = 30\) (number of trials, i.e., number of peas in a group) and \(p=0.75\) (probability of a pea having green pods).

$$ \mu=np = 30\times0.75=22.5 $$

Step2: Calculate the standard deviation

The standard deviation formula for a binomial distribution is \(\sigma=\sqrt{np(1 - p)}\). Substitute \(n = 30\) and \(p = 0.75\) into the formula. First, \(1-p=1 - 0.75=0.25\). Then

$$ \sigma=\sqrt{30\times0.75\times0.25}=\sqrt{5.625}\approx2.4 $$

Answer:

The value of the mean is \(\mu = 22.5\) peas. The value of the standard deviation is \(\sigma\approx2.4\) peas.