With regards to a standard normal distribution complete the following: (a) Find

Question

With regards to a standard normal distribution complete the following: (a) Find P(z < 0), the proportion of the standard normal distribution below the z-score of 0. (b) Find P(z > 0.5), the proportion of the standard normal distribution above the z-score of 1.5. (c) Find P(-1.5 < z < 2.5). (d) Find P( z < 0.75). (e) Find the z-score that separates the lower 80% of standarized scores from the top 20% . . . that is find the z-score corresponding to P80, the 90th percentile value. With regards to a standard normal distribution complete the following: (a) Find P(z < 0), the proportion of the standard normal distribution below the z-score of 0. (b) Find P(z > 0.5), the proportion of the standard normal distribution above the z-score of 1.5. (c) Find P(-1.5 < z < 2.5). (d) Find P( z < 0.75). (e) Find the z-score that separates the lower 80% of standarized scores from the top 20% . . . that is find the z-score corresponding to P80, the 90th percentile value.

Explanation / Answer

a)

Using a table/technology, the left tailed area of this is          
          
P(z <   0   ) =    0.5 [ANSWER]

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b)

Using a table/technology, the left tailed area of this is          
          
P(z <   0.5   ) =    0.691462461
          
Thus, the right tailed area is the complement,          
P(z >   0.5   ) =    0.308537539 [ANSWER]

[Please check this part for typos, as there is a contradiction. If you must use 1.5, here is the alternative:

Using a table/technology, the left tailed area of this is          
          
P(z <   1.5   ) =    0.933192799
          
Thus, the right tailed area is the complement,          
P(z >   1.5   ) =    0.066807201 [ANSWER] ]

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c)

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   [ANSWER]

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d)

Using a table/technology, the left tailed area of this is          
          
P(z <   0.75   ) =    0.773372648 [ANSWER]

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e)

We get the z score from the given left tailed area. As      
      
Left tailed area =    0.8  
      
Then, using table or technology,      
      
z =    0.841621234   [ANSWER]

  

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