Hi, 

variable X with a normal distribution with mean 36 inches and standard deviation 2 inches       

finding the X value approximating the 50th percentile of this distribution

*Note: 

 NORMSINV(0.50)= 0    | area illustrated to the left of center - the mean value

   z= (X - 36)/2 = 0   

       X = 36, the mean value 

 

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