assume that hybridization experiments are conducted with peas having the property that for offspring, there…
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.\na. find the mean and the standard deviation for the numbers of peas with green pods in the groups of 28.\nthe value of the mean is $mu = square$ peas\n(type an integer or a decimal. do not round.)
Answer
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