Watch the video and then complete parts (a) through (g) below.
Click here to watch the video.
Use the frequency distribution to the right to complete parts (a) through $(\mathrm{g})$.
\begin{tabular}{|c|c|}
\hline $\mathbf{x}$ & $\mathbf{f}$ \\
\hline 10 & 2 \\
\hline 14 & 5 \\
\hline 15 & 6 \\
\hline 20 & 5 \\
\hline 25 & 2 \\
\hline
\end{tabular}
(e) Which of the following calculations would result in the mean of the data set and of the frequency distribution? Select all that apply.
A. $\frac{10+14+15+20+25}{5}$
B. $\frac{2(10)+5(14)+6(15)+5(20)+2(25)}{5}$
C. $\frac{2(10)+5(14)+6(15)+5(20)+2(25)}{20}$
D. $\frac{10+10+14+14+14+14+14+15+15+15+15+15+15+20+20+20+20+20+25+25}{20}$
E. $\frac{10+14+15+20+25}{20}$
F. $\frac{2+5+6+5+2}{5}$

Step 1 ：Given the frequency distribution table, we have the values as \(x = [10, 14, 15, 20, 25]\) and their corresponding frequencies as \(f = [2, 5, 6, 5, 2]\).

Step 2 ：The total number of values, which is the sum of the frequencies, is 20.

Step 3 ：The mean of a frequency distribution is calculated by multiplying each value by its frequency, summing these products, and then dividing by the total number of values. So, we calculate the sum of the products as \(2(10)+5(14)+6(15)+5(20)+2(25) = 330\).

Step 4 ：Then, we calculate the mean as \(\frac{330}{20} = 16.5\).

Step 5 ：Now, we need to identify which of the given options would result in this value. We calculate the results for each option: A. \(\frac{10+14+15+20+25}{5} = 16.8\), B. \(\frac{2(10)+5(14)+6(15)+5(20)+2(25)}{5} = 66.0\), C. \(\frac{2(10)+5(14)+6(15)+5(20)+2(25)}{20} = 16.5\), D. \(\frac{10+10+14+14+14+14+14+15+15+15+15+15+15+20+20+20+20+20+25+25}{20} = 16.5\), E. \(\frac{10+14+15+20+25}{20} = 4.2\), F. \(\frac{2+5+6+5+2}{5} = 4.0\).

Step 6 ：From the above calculations, we can see that only options C and D result in the mean of the data set and of the frequency distribution.

Step 7 ：Final Answer: The calculations that would result in the mean of the data set and of the frequency distribution are options C and D. Therefore, the correct answer is \(\boxed{['C', 'D']}\).