Answer:
See Explanation
Step-by-step explanation:
The question is incomplete, as a table or chart that shows the required data is not given.
To answer this, I will make use of the following frequency table
[tex]\begin{array}{cccccc}{Miles} & {1} & {1\frac{1}{2}} & {2} & {2\frac{1}{2}} & {3} \ \\ {Students} & {10} & {0} & {5} & {10} & {5} \ \end{array}[/tex]
From the above table.
[tex]2\frac{1}{2}\ miles \to 10\ students[/tex]
[tex]3\ miles \to 5\ students[/tex]
So:
[tex]2\frac{1}{2}\ or\ 3\ miles = 10+5[/tex]
[tex]2\frac{1}{2}\ or\ 3\ miles = 15\ students\\[/tex]
From the above table.
[tex]1\ miles \to 10\ students[/tex]
[tex]1\frac{1}{2}\ miles \to 0\ students[/tex]
So:
[tex]1\frac{1}{2}\ or\ 1\ miles = 0+10[/tex]
[tex]1\frac{1}{2}\ or\ 1\ miles = 10\ students[/tex]
The difference (d) is then calculated by subtracting the number of students in both categories
[tex]d = 15\ students - 10\ students[/tex]
[tex]d = 5\ students[/tex]
