Answer:
a) The mean time it took the students, t ≈ 17.08 minutes
b) The fraction of the students that finished under 20 minutes is 2/3 of the students
Step-by-step explanation:
a) The mean of the frequency of grouped data is given as follows;
The given data arranged in tabular form is presented as follows;
[tex]\begin{array}{cccc}Time \, (t) \, in \, minutes & Midpoint, m & Frequency, f & Product \ m \cdot f \\5\leq t < 10 & 7.5 & 2 & 15 \\10\leq t < 15 & 12.5 & 9 & 112.5\\15\leq t < 20 & 17.5 & 5 & 87.5 \\20\leq t < 25 & 22.5 & 5 & 112.5 \\25 \leq t < 30 & 27.5 & 3 & 82.5 \\Sum & 87.5 & N = 24 & \Sigma (m\cdot f) = 410 \end{array}[/tex]
Therefore, the mean time it took the students = ∑(f·m)/N = 410/24 = [tex]17\frac{1}{12}[/tex]
The mean time it took the students, t ≈ 17.08 minutes
b) The number of the students that finished under 20 minutes = 2 + 9 + 5 = 16 students
The fraction of the students that finished under 20 minutes = 16/24 of the = 2/3 of the students.
