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 - μ)/$4 = .5244

  $14 - 4(.5244) = μ 

    $11.90 = μ    

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