Suppose your statistics instructor gave six examinations during the semester. You received the following grades (percent correct): 79, 64, 84, 82,92, and 77. Instead of averaging the six scores, the instructor indicated he would randomly select two grades and report that grade to the student records office.
a. How many different samples of two test grades are possible?
b. List all possible samples of size two and compute the mean of each.
c. Compute the mean of the sample means and compare it to the population mean.

a. 15

b.

Sr.no Samples Sample mean

1           (79,64)         71.5

2          (79,84)          81.5

3          (79,82)          80.5

4          (79,92)          85.5

5          (79,77)           78

6         (64,84)            74

7          (64,82)            73

8          (64,92)            78

9          (64,77)            70.5

10        (84,82)            83

11         (84,92)            88

12        (84,77)             80.5

13        (82,92)            87

14        (82,77)            79.5

15        (92,77)            84.5

c.

mean of sample mean=population mean=79.67

Step-by-step explanation:

a.

The different samples of two test grade are nCr, where n=6 and r=2.

nCr=6C2=6!/2!(6-2)!=6*5*4!/2!4!=30/2=15.

Thus, there are 15 different samples of two test grade.

b.

All possible samples are listed below:

Sr.no Samples

1           (79,64)        

2          (79,84)          

3          (79,82)          

4          (79,92)        

5          (79,77)          

6         (64,84)            

7          (64,82)

8          (64,92)

9          (64,77)

10        (84,82)

11         (84,92)

12        (84,77)

13        (82,92)

14        (82,77)

15        (92,77)

The sample means for each sample can be calculated as

Sr.no Samples Sample mean

1           (79,64)         (79+64)/2=71.5

2          (79,84)          (79+84)/2=81.5

3          (79,82)          (79+82)/2=80.5

4          (79,92)          (79+92)/2=85.5

5          (79,77)           (79+77)/2=78

6         (64,84)            (64+84)/2=74

7          (64,82)            (64+82)/2=73

8          (64,92)            (64+92)/2=78

9          (64,77)            (64+77)/2=70.5

10        (84,82)            (84+82)/2=83

11         (84,92)            (84+92)/2=88

12        (84,77)             (84+77)/2=80.5

13        (82,92)            (82+92)/2=87

14        (82,77)            (82+77)/2=79.5

15        (92,77)            (92+77)/2=84.5

c.

The sample means of sample mean μxbar will calculated by taking average of sample means

μxbar=(71.5+ 81.5+ 80.5+ 85.5+ 78+ 74+ 73+ 78+ 70.5+ 83+ 88+ 80.5+ 87+ 79.5+ 84.5)/15

μxbar=1195/15=79.67

Population mean=μ=(79+64+84+82+92+77)/6

μ=478/6=79.67

Sample means of sample mean μxbar=Population mean μ.

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: