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 28. 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 28.
the value of the mean is $mu = square$ peas
(type an integer or a decimal. do not round.)

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem where $n = 28$ (number of trials) and $p=0.75$ (probability of success).

Step2: Calculate the mean

The formula for the mean of a binomial distribution is $\mu = np$. So, $\mu=28\times0.75$.
$\mu = 21$

Step3: Calculate the standard - deviation

The formula for the standard deviation of a binomial distribution is $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 28$ and $p = 0.75$. First, find $1-p=1 - 0.75=0.25$. Then $np(1 - p)=28\times0.75\times0.25 = 5.25$. So, $\sigma=\sqrt{5.25}\approx2.2913$. But we are only asked for the mean here.

Answer:

21