Question 429444
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Hi
hourly rates have a normal distribution with a standard deviation of $4.00
30 percent of all part-time seasonal employees make more than $14.00 an hour
 P(all part-time seasonal employees make more than $14.00 an hour) = .30
then correspondingly  z = .5244
  
 ($14 - &#956;)/$4 = .5244
  $14 - 4(.5244) = &#956; 
    $11.90 = &#956;