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=2.2 \\) peas.
(round to one decimal place as needed.)
b. use the range rule of thumb to find the values separating results that are significantly low or significantly high.
values of \\( \square \\) peas or fewer are significantly low.
(round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean and standard deviation for part a

Given \(n = 26\), \(p=0.75\), \(q = 1 - p=0.25\)
The mean formula for a binomial distribution is \(\mu=np\).
Substitute \(n = 26\) and \(p = 0.75\) into the formula: \(\mu=26\times0.75 = 19.5\)
The standard deviation formula for a binomial distribution is \(\sigma=\sqrt{npq}\)
Substitute \(n = 26\), \(p = 0.75\) and \(q = 0.25\) into the formula: \(\sigma=\sqrt{26\times0.75\times0.25}=\sqrt{4.875}\approx2.2\)

Step2: Use the range rule of thumb for part b

The range rule of thumb states that values are significantly low if \(x\leq\mu - 2\sigma\) and significantly high if \(x\geq\mu + 2\sigma\)
First, calculate \(\mu - 2\sigma\): \(19.5-2\times2.2=19.5 - 4.4 = 15.1\)
Then, calculate \(\mu + 2\sigma\): \(19.5+2\times2.2=19.5 + 4.4 = 23.9\)

Answer:

a. Mean \(\mu = 19.5\) peas, standard deviation \(\sigma\approx2.2\) peas
b. Significantly low: \(15.1\) peas or fewer; significantly high: \(23.9\) peas or more