From short math tests during the first semester, Michal received the following grades consecutively: $1,\,3,\,4,\,1,\,2,\,3,\,2,\,1,\,1,\,3,\,3,\,4$. His classmates helped him calculate statistical measures from this set of grades.
First, they created a frequency table of the grades:
\begin{array}{|l|c|c|c|c|c|} \hline Grade&1&2&3&4&5\cr \hline Frequency&4&2&4&2&0\cr \hline \end{array}
(1) Petr calculated the arithmetic mean. He had to add up all the grades and divide them by their total count. He used the frequency table: $$(1\cdot4+2\cdot2+3\cdot4+4\cdot2+5\cdot0)∶(4+2+4+2+0)=28∶12\approx2.33$$
(2) Martina was tasked with determining the mode of the dataset. She claimed that the mode is the value of the grade that appears most frequently in the given set. Upon inspecting the table, she discovered that Michal received the same count of ones and threes. According to Martina, Michal's grades have two modes:
$$\mbox{Mode}_1=1,\quad\mbox{Mode}_2=3$$
(3) Jirka sought the median of the dataset. Jirka remembered that the median is the middle value. He arranged all the grades from best to worst: $$1,\,1,\,1,\,1,\,2,\,2,\,3,\,3,\,3,\,3,\,4,\,4$$
Then, he looked for the grade that is “in the middle”. Knowing there were $12$ grades (an even count), there were two values in the middle. Jirka designated both of these grades as the median: $$\mbox{Med}_1=2,\quad\mbox{Med}_2=3$$
Did they calculate the statistical measures correctly? Did anyone make a mistake in the calculations? Choose the true statement:
Petr made a mistake. He should have added the values of the grades $(1+2+3+4+5)$ from the frequency table and divided them by the number of possible grades. The arithmetic mean should be: $$(1+2+3+4+5):5=3$$
Martina made a mistake. A dataset cannot have more than one mode. If two values have the same frequency, the mode is the arithmetic mean of these values, i.e., $$\mbox{Mode}=(1+3):2=2.$$
Jirka made a mistake. The median cannot simultaneously be two values. With an even count of values, the median is the arithmetic mean of the two middle values, i.e., $\mbox{Med}=(2+3):2=2.5.$
All classmates proceeded correctly and made no mistakes.