Given that the population mean is 190 and the standard devia

Given that the population mean is 190 and the standard deviation is 36 on a continuous normally distributed scale find find P(140<X<194).

So I did all of the conversions and I got

P(-1.39<Z<0.111) <--- the solution manual agrees with this.

0.9177-0.548=0.4177, <--- the solution manual says otherwise (?)

but the solution in the book says 0.4615. What\'s throwing me off is that for P(Z<-1.39) it gave a value of 0.0823. How did they get that? My table doesn\'t go that far and I can\'t figure out where it came from. Any help would be greatly appreciated.

Solution

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    140      
x2 = upper bound =    194      
u = mean =    190      
          
s = standard deviation =    36      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.39      
z2 = upper z score = (x2 - u) / s =    0.11      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.0823      
P(z < z2) =    0.5438      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.4615      
*********************

Please use the z table which gives out the LEFT TAILED AREAS for the z scores.

It seems you are getting right tailed areas sometimes (0.9177), sometimes left tailed areas (0.5438).

You may notice that 0.9177 + 0.0823 = 1. You have the right tailed area here, instead of a left tailed area. I think this is the problem.