11) A genetics experiment involves a population of fruit flies

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11) A genetics experiment involves a population of fruit flies consisting of 1 male named
Mike and 3 females named Anna, Barbara, and Chris. Assume that two fruit flies are
randomly selected with replacement.
a) After listing the 16 different possible samples, find the proportion of females in
each sample, then use a table to describe the sampling distribution of the proportions
of females.
SAMPLES (Make a chart of the possible combinations. I filled in a couple
to get you started…)
M, M
M, B
M, A
M, C
A, M
Mike/Mike
means no
females
A, B
Mike/Barb means
½ are females:
50% = 0.5
B, A
C, M
PROPORTION OF FEMALES (Each proportion is the fraction of females in
that sample.)
0
0.5
0.5
0.5
0.5
1.0
1.0
Barb/Anna means
all are females:
100% = 1.0
SAMPLING DISTRIBUTION OF SAMPLE MEANS (After you fill in the
charts, count up the number of times that each proportion (0, 0.5, and 1)
occurs and make it into a fraction out of 16. 16 comes from the number of
possible samples.)
Proportion
Probability
0.5
0
1/16 = 0.0625
0.5
1.0
b) Find the mean of the sampling distribution.
  x  Px  (0  0.0625)  (0.5  ____)  (1.0  ____)  ?
c) Is the mean of the sampling distribution equal to the mean of the population of five
listed values? Are those means always equal?
To answer this question, read through pages 270-274.
I hope this helps… 
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