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 20. Find the mean and the standard deviation for the numbers of peas with green pods in the groups of 20.

Answer 1

Answer : Give data

p=0.75 and q=(1- p) =(1- 0.75) =0.25

n=20

Here X~B(n, p) , That is X~B(20, 0.75)

Find the mean and variance

In binomial distribution

mean = (np) and

standard deviation= sqrt(npq)

Mean = (np) = 20*0.75 = 15

Mean = 15

Standard deviation =sqrt(npq) =sqrt(20*0.75*0.25)

=sqrt (3.75)

Standard deviation =1.9364

Ans : mean =15 and standard deviation =1.9364