Question 471495
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Hi,
a mean of 24 burgers an hour
between 24 and 42 burgers 49.85% of the time
Note: %99.7 of a normal distribution lies within 
3 standard deviations from the mean:  99.7%/2 = 49.85% represents the right side
  {{{z = blue(x - mu)/blue(sigma)}}} = 3
|standard of deviation is 6: as  
  z = (42-24)/{{{highlight(6)}}} = 3

With a mean of 24 burgers an hour and sd = 6  
z = (12-24)/6 = -2 and z  = (18-24)/6 = -1
P(12 to 18) = P ( -2 < z < -1) = .158655 - .02275 = .136