A genetics experiment involves a population of fruit flies consisting of 2 males named Alfonso and Bart and 2 females named Carla and Diane. Assume that two fruit flies are randomly selected with replacement.
(a) Find the mean of the sampling distribution.
(b) Is the mean of the sampling distribution [from part (a)] equal to the population proportion of females?

Step-by-step explanation:

Mean is given by

∑xp

Here, x is value of X and p is probability associated with x.

a. Let, A, B, C, and D represent Alfonso, Bart, Carla, and Diane respectively.

The Possible samples with proportion of female (in brackets) are

AA(0),

AB(0),

AC(0.5),

AD(0.5),

BB(0),

BA(0),

BC(0.5),

BD(0.5),

CA(0.5),

CB(0.5),

CC(1),

CD(1),

DA(0.5),

DB(0.5),

DC(1),

DD(1).

Clearly, from the above sample proportion of females are 0,0.5,1.

a. Proportion of females (0)

Then, the probability Is 4/16=¼=0.25

Proportion of females (1)

Then, the probability Is 8/16=½=0.5

Proportion of females (2)

Then, the probability Is 4/16=¼=0.25

b. The mean proportion is give as

Mean=

∑xp=0*0.25+0.5*0.5+1*0.25=0.5

Therefore,

Mean=0.5

Population proportion of female=2/4=0.5

Clearly, the mean of the probability distribution is equal to the population proportion of females.