Mr. Sanders teaches a painting class. He makes the histogram below to summarize the
number of paintings each student has completed since the class began. Mr. Sanders says
that the histogram shows that of the students have finished 12 or more paintings.
Number of Paintings
Finished by Students
Number of Students
N
0-2 3-5 6-8 9-11 12-14 15-17
Number of Finished Paintings
What error did Mr. Sanders likely make, and what is the actual fraction of students who have
completed 12 or more paintings? Use the drop-down menus to explain his reasoning and correct
his mistake.
Mr. Sanders likely found that of the students have completed 12 or more paintings because
there are Choose... bars on the histogram, and choose... of the bars represent
students with 12 or more completed paintings. The actual fraction of the students who have
completed 12 or more paintings is Choose...

Mr. Sanders most certainly discovered that one-third of pupils have done 12 or more paintings because the histogram has six bars, two of which reflect students who have completed 12 or more paintings. The actual percentage of pupils who have completed 12 or more paintings is [tex]\frac{4}{19}[/tex].

Mr. Sanders' class has 19 students. That histogram has six bars. He most likely assumed it because [tex]\frac{2}{6} = \frac{1}{3}, \frac{1}{3}[/tex] of his pupils finished 12 or more paintings.
That is not the case. 2 different isn't even a whole number. The other dates are[tex] \frac{2}{3}, \frac{4}{19}, \ and \ \frac{4}{17}[/tex]. Two-thirds and 4 are not whole figures either.
That leaves [tex]\frac{4}{19}[/tex]. Four-nineteenths equal a complete number four. Since there are no such things as partial pupils, we must enter [tex]\frac{4}{19}[/tex].

Therefore, the answer is "[tex]\to (1^{st}\ box) = 6 (2^{nd}\ box) = 2 (3^{rd}\ box) = \frac{2}{3}[/tex]"

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