Given that z is a standard normal random variable compute th
Solution
a)
 Using a table/technology, the left tailed area of this is          
           
 P(z <   -1   ) =    0.158655254 [ANSWER]
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b)
Using a table/technology, the right tailed area of this is          
           
 P(z >   -1   ) =    0.841344746 [ANSWER]
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c)
Using a table/technology, the right tailed area of this is          
           
 P(z >   -1.5   ) =    0.933192799 [ANSWER]
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d)
Using a table/technology, the right tailed area of this is          
           
 P(z >   -2.5   ) =    0.993790335 [ANSWER]
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e)
z1 = lower z score =    -3      
 z2 = upper z score =     0      
           
 Using table/technology, the left tailed areas between these z scores is          
           
 P(z < z1) =    0.001349898      
 P(z < z2) =    0.5      
           
 Thus, the area between them, by subtracting these areas, is          
           
 P(z1 < z < z2) =    0.498650102   [ANSWER]  

