Let z denote a random variable having a normal distribution

Let z denote a random variable having a normal distribution with = 0 and = 1. Determine each of the probabilities below. (Round all answers to four decimal places.)

(a) P(z < 0.1) =  

(b) P(z < -0.1) =  

(c) P(0.40 < z < 0.84) =  

(d) P(-0.84 < z < -0.40) =  

(e) P(-0.40 < z < 0.84) =  

(f) P(z > -1.25) =  

(g) P(z < -1.5 or z > 2.50) =

Solution

a)

Using a table/technology, the left tailed area of this is          
          
P(z <   0.1   ) =    0.539827837 [ANSWER]

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b)

Using a table/technology, the left tailed area of this is          
          
P(z <   -0.1   ) =    0.460172163 [ANSWER]

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c)

z1 = lower z score =    0.4      
z2 = upper z score =     0.84      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.655421742      
P(z < z2) =    0.799545807      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.144124065   [ANSWER]

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d)

z1 = lower z score =    -0.84      
z2 = upper z score =     -0.4      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.200454193      
P(z < z2) =    0.344578258      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.144124065   [ANSWER]

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e)

z1 = lower z score =    -0.4      
z2 = upper z score =     0.84      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.344578258      
P(z < z2) =    0.799545807      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.454967548   [ANSWER]

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f)

Using a table/technology, the left tailed area of this is          
          
P(z <   -1.25   ) =    0.105649774
          
Thus, the right tailed area is the complement,          
P(z >   -1.25   ) =    0.894350226 [ANSWER]

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g)

z1 = lower z score =    -1.5      
z2 = upper z score =     2.5      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.066807201      
P(z < z2) =    0.993790335      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.926983133      

Thus, those outside this interval is the complement =    0.073016867   [ANSWER]