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 μ = 21 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: Recall binomial distribution formula for mean

For a binomial distribution $X\sim B(n,p)$, the mean $\mu = np$. Here $n = 28$ (number of trials/sample size) and $p=0.75$ (probability of success). So $\mu=np = 28\times0.75=21$.

Step2: Recall binomial distribution formula for standard - deviation

The standard deviation $\sigma=\sqrt{np(1 - p)}$. Substitute $n = 28$ and $p = 0.75$. First, calculate $1-p=1 - 0.75 = 0.25$. Then $np(1 - p)=28\times0.75\times0.25=5.25$. So $\sigma=\sqrt{5.25}\approx2.3$.

Answer:

The value of the standard deviation is $\sigma = 2.3$ peas.