#18. suppose that the scores of architects on a particular creativity test are normally ddistributed. using a normal curve table, what percentage of architects have Z scores (a)above .10, (b)below .10, (c)above .20, (d) below .20, (e) above 1.10, (f) below 1.10, (g) above -.10, and (h) below _.10 -in the example in problem 18, using a noral curve table, what is the minimum Z sscore an architect can have onthe creativity test to be in the (a) top %50%, (b) top 40%, (c)top 60%, (d)top 30%, and (e)top20%

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Asked Sep 23, 2019

#18. suppose that the scores of architects on a particular creativity test are normally ddistributed. using a normal curve table, what percentage of architects have Z scores (a)above .10, (b)below .10, (c)above .20, (d) below .20, (e) above 1.10, (f) below 1.10, (g) above -.10, and (h) below _.10 -in the example in problem 18, using a noral curve table, what is the minimum Z sscore an architect can have onthe creativity test to be in the (a) top %50%, (b) top 40%, (c)top 60%, (d)top 30%, and (e)top20%

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Step 1

Hey, since there is multiple sub part questions posted, we will answer first three sub part questions. If you want any specific sub part question to be answered then please submit that sub part questions only or specify the question number in your message.

Step 2

(18.a) Obtain the percentage of architects have Z scores above 0.10:     

The P(Z>0.10)=1–P(Z<0.10)

The percentage of architects have Z scores above 0.10 is obtained below:

Procedure for finding the probability value (confidence level) from standard normal table:

The z value be 0.10

  1. In the standard normal table first locate the value 0.1 in the first z column.
  2. Locate the value of 0.00 in the first z row.
  3. Move right until the column of 0.1 is reached.
  4. From the located 0.1 column move down until the row 0.10 is reached.
  5. Locate the probability value, by the intersection of the row and column values gives the area to the left of z.

The area to the left of Z=0.10 is 0.5398.

The required percentage is,

P(Z>0.10)1-P(Z<0.10)
1-0.5398
0.4602
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P(Z>0.10)1-P(Z<0.10) 1-0.5398 0.4602

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Step 3

(18.b) Obtain the percentage of architects have Z scores below 0.10:     

The percentage of architects have Z scores below 0.10 is obtained below:

Procedure for finding the probability value (confidence level) from standard normal table:

The z value be 0.10

  1. In the standard normal table first locate the value 0.1 in the first z column.
  2. Locate the value of 0.00 in the first z row.
  3. Move right until the column of 0.1 is reached.
  4. From the located 0.1 column move down until the row 0.10 is reached.
  5. Locate t...

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