Answer:
The margin of error is  [tex]ME = 2.2962[/tex]
Step-by-step explanation:
From the question we are told that
   The number of student is [tex]n = 500[/tex]
   The highest amount is  [tex]A =[/tex]$200
   The lowest amount is  [tex]B =[/tex] $75
    The sample mean is  [tex]x =[/tex] &140
The Standard deviation  of this set is mathematically evaluated as
      [tex]s = \frac{A-B}{4}[/tex]
Substituting values
      [tex]s = \frac{200-75}{4}[/tex]
      [tex]s = 31.25[/tex]
The margin of error (ME) is mathematically evaluated as
      [tex]ME = t_{n-1}__{\alpha }} * \frac{s}{\sqrt{n} }[/tex]
Where  [tex]t_{n-1}__{\alpha }}[/tex] is the critical value for [tex]\alpha = 0.05[/tex] i.e the significance level
   From the critical value table this is  [tex]t_{n-1}__{\alpha }} = 1.649[/tex]
So
    [tex]ME = 1.649 * \frac{31.25}{\sqrt{500} }[/tex]
   [tex]ME = 2.2962[/tex]
