Question

Suppose that the score of architects on a particular creativity test are normally distributed. Using a normal curve table, what percentage of architects have a z scores above .10, below .10, above .20, below .20, above 1.10, below 1.10, above -.10 and below -.10

the example problem above, using the normal curve table, what is the minimum z score an architect can have on the creativity test to be in the top 50%, top 40%, top 60%, top 30% and top 20% please show the formula used to get the answer

Answer 1

Solution:

Given that,

Using standard normal table,

a ) P ( Z > 0.10 )

1 - P ( Z < 0.10 )

= 1 - 0.5398

= 0.4602

Probability = 46.02%

b ) P ( Z < 0.10 )

= 0.5398

Probability = 53.98%

c ) P ( Z > 0.20 )

1 - P ( Z < 0.20 )

= 1 - 0.5793

= 0.4207

Probability = 42.07%

d ) P ( Z < 0.20 )

= 0.5793

Probability = 57.93%

f ) P ( Z > 1.10 )

1 - P ( Z < 1.10 )

= 1 - 0.8643

= 0.1357

Probability = 13.57%

e) P ( Z < 1.10 )

= 0.8643

Probability = 86.43%

f ) P ( Z > - 0.10 )

1 - P ( Z < - 0.10 )

= 1 - 0.4602

= 0.5398

Probability = 53.98%

g ) P ( Z < - 0.10 )

= 0.4602

Probability = 46.02.%

h ) P ( Z > z ) = 50%

1 - P ( Z <  z ) = 0.50

P ( Z <  z ) = 1 - 0.50

P ( Z < 0 ) = 0.50

z = 0

i ) P ( Z > z ) = 40%

1 - P ( Z <  z ) = 0.40

P ( Z <  z ) = 1 - 0.40

P ( Z < 0.25 ) = 0.60

z = 0.25

j ) P ( Z > z ) = 60%

1 - P ( Z <  z ) = 0.60

P ( Z <  z ) = 1 - 0.40

P ( Z < - 0.25 ) = 0.40

z = - 0.25

k ) P ( Z > z ) = 30%

1 - P ( Z <  z ) = 0.30

P ( Z <  z ) = 1 - 0.30

P ( Z < 0.52 ) = 0.70

z = 0.52

l ) P ( Z > z ) = 20%

1 - P ( Z <  z ) = 0.20

P ( Z <  z ) = 1 - 0.20

P ( Z < 0.84 ) = 0.80

z = 0.84