Use a standard normal table to obtain the areas under the no
Solution
a.
It is graph OPTION A, by inspection.
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Thus, the two z scores are          
           
 z1 = lower z score =    -1.75      
 z2 = upper z score =    1.34      
           
 Using table/technology, the left tailed areas between these z scores is          
           
 P(z < z1) =    0.040059157      
 P(z < z2) =    0.909877328      
           
 Thus, the area between them, by subtracting these areas, is          
           
 P(z1 < z < z2) =    0.869818171      
           
 Thus, those outside this interval is =    0.1302 [ANSWER]
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b.
It is graph option A, by inspection.
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Thus, the two z scores are          
           
 z1 = lower z score =    0.64      
 z2 = upper z score =    1.69      
           
 Using table/technology, the left tailed areas between these z scores is          
           
 P(z < z1) =    0.7389137      
 P(z < z2) =    0.954486023      
           
 Thus, the area between them, by subtracting these areas, is          
           
 P(z1 < z < z2) =    0.215572322      
           
 Thus, those outside this interval is =    0.7844 [ANSWER]      

