Consider the following sample data. Sample A: 4, 6, 8 Sample B: 63, 65, 67 Sample C: 1,010; 1,012; 1,014 (a) Find the mean and standard deviation for each sample. Sample A: Sample B: Sample C: Mean 6 6 Correct 65 65 Correct 1,002 1,002 Incorrect Sample Standard Deviation 3 3 Incorrect 3 3 Incorrect 3 3 Incorrect (b) What does this exercise show about the standard deviation? The idea is to illustrate that the standard deviation is not a function of the value of the mean. The idea is to illustrate that the standard deviation is a function of the value of the mean.

(a) Sample A: mean = 6, sd = 1.63

Sample B: mean = 65, sd = 1.63

Sample C: mean = 1,012, sd = 1.63

(b) The idea is to illustrate that the standard deviation is a function of the value of the mean.

Step-by-step explanation:

(a) Sample A: 4, 6, 8

Mean = (4+6+8)/3 = 18/3 = 6

Standard deviation = sqrt[((4-6)^2 + (6-6)^2 + (8-6)^2) ÷ 3] = sqrt(8 ÷ 3) = sqrt(2.667) = 1.63

Sample B: 63, 65, 67

Mean = (63+65+67)/3 = 195/3 = 65

Standard deviation = sqrt[((63-65)^2 + (65-65)^2 + (67-65)^2) ÷ 3] = sqrt(8 ÷ 3) = sqrt(2.667) = 1.63

Sample C: 1,010, 1,012, 1,014

Mean = (1,010+1,012+1,014)/3 = 3,036/3 = 1,012

Standard deviation = sqrt[((1,010-1012)^2 + (1,012-1,012)^2 + (1,014-1,012)^2) ÷ 3] = sqrt(8 ÷ 3) = sqrt(2.667) = 1.63

(b) The exercise shows that the standard deviation is a function of the value of the mean.

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1