Question

what conclusion can be derived by comparing the central tendencies of the two data sets? a: {7, 6, 3, 1, 6, 2, 4, 6, 3, 5} b: {2, 2, 2, 3, 4, 5, 2, 8, 7, 6} a. the mean of set a is smaller than the mean of set b. b. the median of set a is greater than the median of set b. c. the median and the mean of set b are greater than those of set a. d. the mode of set b is greater than the mode of set a

Explanation:

Step1: Calculate mean of set A

The mean formula is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For set A = {7, 6, 3, 1, 6, 2, 4, 6, 3, 5}, $n = 10$ and $\sum_{i=1}^{10}x_{i}=7 + 6+3 + 1+6+2+4+6+3+5=43$. So the mean of set A, $\bar{x}_A=\frac{43}{10}=4.3$.

Step2: Calculate mean of set B

For set B = {2, 2, 2, 3, 4, 5, 2, 8, 7, 6}, $n = 10$ and $\sum_{i = 1}^{10}x_{i}=2+2+2+3+4+5+2+8+7+6 = 41$. So the mean of set B, $\bar{x}_B=\frac{41}{10}=4.1$.

Step3: Arrange set A in ascending - order

Set A in ascending - order: {1, 2, 3, 3, 4, 5, 6, 6, 6, 7}. Since $n = 10$ (even), the median is the average of the 5th and 6th values. Median of set A, $M_A=\frac{4 + 5}{2}=4.5$.

Step4: Arrange set B in ascending - order

Set B in ascending - order: {2, 2, 2, 2, 3, 4, 5, 6, 7, 8}. Since $n = 10$ (even), the median is the average of the 5th and 6th values. Median of set B, $M_B=\frac{3 + 4}{2}=3.5$.

Step5: Find mode of set A

The mode of set A is 6 (it appears 3 times).

Step6: Find mode of set B

The mode of set B is 2 (it appears 4 times).

Answer:

B. The median of set A is greater than the median of set B.