Explanation
we will begin by finding the mean and median of the data set
The mean is simply the average of the set, which will be
[tex]mean=\frac{5+0+5+2+0+10+7+8+10+21+5+8+2+5+3+5}{16}=\frac{96}{16}=6[/tex]The median is
[tex]\begin{gathered} \mathrm{The\:median\:is\:the\:value\:separating\:the\:higher\:half\:of\:the\:data\:set,\:from\:the\:lower\:half.} \\ \mathrm{If\:the\:number\:of\:terms\:is\:odd,\:then\:the\:median\:is\:the\:middle\:element\:of\:the\:sorted\:set} \\ If\:the\:number\:of\:terms\:\:is\:even,\:then\:the\:median\:is\:the\:arithmetic\:mean\:of\:the\:two\:middle\:elements\:of\:the\:sorted\:set \end{gathered}[/tex]Thus, we have the median as 5
To check which is a better measure, we will have to check the skewness
The skew value is 1.51
This means it is positively skewed
Thus
If the distribution is positively skewed then the mean is greater than the median which is in turn greater than the mode.
Therefore, the answer is