Two distributions of data are being analyzed. Distribution A has a mean of 500 a

Question

Two distributions of data are being analyzed. Distribution A has a mean of 500 and a standard devation equal to 8.0. Based on this information, use the coefficient of variation Choose the correct answer below deviation equal to 200. Distribution B has a mean of 40 and a standa to determine which distribution has greater relative vaniation A· Distribution A has greater relative vanation since its coefficient of variation is less than distribution Bs O B. Distribution B has greater relative vanation since its coefficient of vanation is greater than distribution A. O c. Distnbution B has greater relative variation since its coefficient of variation is less than distribution A's O D. Distibution A has greater relative variation since its coeffcient of variation is greater than distibution B's. Click to select your answer

Explanation / Answer

Q1)

Coefficient of variation = (std deviation/mean) *100us

Thus, Coefficient of variation for A = (200/500)*100 = 40

Coefficient of variation for B = (8/40)*100 = 20

Thus, option D

Q2)

Mean of 2004 data = (0.4+1.5-0.1+0.4+1.6+0.5+1.5-0.4-1.4)/9 =4/9 = 0.444

Variance = sum((x-mean)^2) / n = (0.16+..+1.96)/9 = 1.084

std deviation of 2004 = sqrt(Variance) = 1.041

Interquartile range = Q3-Q1

Q1 is the "middle" value in the first half of the rank-ordered data set

Q3 is the "middle" value in the second half of the rank-ordered data set

Ordering in ascending order

-1.4 -0.4 -0.1 0.4 0.4 0.5 1.5 1.5 1.6

Q1 = taking ratio of 2nd and 3rd = (-0.4-0.1)/2 = -0.25

Q3 =taking ratio of 7th and 8th = 1.5

Thus, IQR for 2004 = 1.5 -(-0.25) = 1.75

Mean of 2005 = (0.6+0,9+2.2+0,9)/4 = 1.15

Variance = sum((x-mean)^2) / n = (0.36+..+0.81)/4 = 1.705

std deviation of 2005 = sqrt(Variance) = 1.306

Interquartile range = Q3-Q1

Q1 is the "middle" value in the first half of the rank-ordered data set

Q3 is the "middle" value in the second half of the rank-ordered data set

rdering in ascending order

0.6 0.9 0.9 2.2

Q1 = (0.6+0.9)/2 = 0.75

Q3 = (0.9+2.2)/2 = 1.55

IQR = 1.55-0.75 = 0.80

Similarly solve next part

Range = difference between max and min value

use same formula for variance and std deviation from previous example

X 0.4 1.5 -0.1 0.4 1.6 0.5 1.5 -0.4 -1.4 X-mean -0.044 1.056 -0.544 -0.044 1.156 0.056 1.056 -0.844 -1.844 (x-mean)^2 0.16 2.25 0.01 0.16 2.56 0.25 2.25 0.16 1.96

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