Customers arrive at a movie theatre at the advertised movie

Customers arrive at a movie theatre at the advertised movie time only to find they have to sit through several previews and prepreview ads before the movie starts.

Solution

Here, s = 4 min = 240 seconds.

A)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    240  
E = margin of error =    75  
      
Thus,      
      
n =    39.33653832  
      
Rounding up,      
      
n =    40   [ANSWER]

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b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    240  
E = margin of error =    60  
      
Thus,      
      
n =    61.46334113  
      
Rounding up,      
      
n =    62   [ANSWER]