Let a population consist of the values 8 cigarettes, 9 cigarettes, and 20 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Mean Absolute Deviation Sample {8,8} {8,9} {8,20} {9,8} {9,9} {9,20} {20,8} {20,9} {20,20} (Type integers or decimals rounded to one decimal place as needed.)

This is part two of a question already asked. I have provided the answer sent, now there is a part 2 to the problem...its shown in picture 2.Part 2 is needed.

3. Solve. x – 4 5 64 x +4 x - 16 O A. {-9} O B. {9} C. {-4, 4} O D. {4}

Question D and E were not answered.