Question 617130
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Hi, 
average income is approximately $45,000/year in this state, with a SD of about $9000.
using a normal distribution to approximate values that {{{89}}}%  of houseolds lie
 1-.89 = .11, .11/2 = .055  Excel NORMSINV(.055)gives z = 1.59819
 1.59819 = (X -45000)/9000 
  X = 1.598*9000 + 45000 = $59,382
and
 -1.59819 = (x -45000)/9000 
x = -1.598*9000 + 45000 = $30,618