Question

the frequency table summarizes the 17 values in data set a. data set b consists of the 17 values from data set a as well as the values 11 and 12. which statement correctly compares the mean and median of data set a to the mean and median of data set b?
a the mean of data set a is equal to the mean of data set b, and the median of data set a is equal to the median of data set b.
b the mean of data set a is equal to the mean of data set b, but the median of data set a is greater than the median of data set b.
c the mean of data set a is greater than the mean of data set b, but the median of data set a is equal to the median of data set b.
d the mean of data set a is greater than the mean of data set b, and the median of data set a is greater than the median of data set b.

Explanation:

Step1: Calculate the mean of data set A

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}f_{i}}{\sum_{i = 1}^{n}f_{i}}\).
For data set A: \(\sum_{i = 1}^{n}x_{i}f_{i}=40\times0 + 41\times4+42\times1+43\times4+44\times0=164 + 42+172=378\), \(\sum_{i = 1}^{n}f_{i}=0 + 4+1+4+0 = 9\). So the mean of data set A is \(\frac{378}{9}=42\).
The data set A (in order): \(41,41,41,41,42,43,43,43,43\). The median (the 5 - th value) is \(42\).

Step2: Calculate the mean of data set B

Data set B has \(9 + 2=11\) values. \(\sum_{i = 1}^{n}x_{i}f_{i}\) for data set B: \(40\times0 + 41\times4+42\times1+43\times4+44\times0+11\times1+12\times1=164 + 42+172+11+12=401\), \(\sum_{i = 1}^{n}f_{i}=11\). The mean of data set B is \(\frac{401}{11}\approx36.45\).
The data set B (in order): \(11,12,41,41,41,41,42,43,43,43,43\). The median (the 6 - th value) is \(42\).

Answer:

C. The mean of data set A is greater than the mean of data set B, but the median of data set A is equal to the median of data set B.