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: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq

so, converting firstly the mixed fractions to improper fractions and then "adding".