Question

An education researcher has been studying test anxiety. She has found that the distribution of her variable follows a normal curve. Using a normal curve table, what percentage of students have Z scores

a. Below 1.2

b. Above 1.2

c. Below -1.2

d. Above -1.2

e. Below .10

f. Above -2.05

g. Above .76

h. Below 1.50

Answer 1

Part a)
P ( Z < 1.2 ) = 0.8849

Percentage = 0.8849 * 100 = 88.49%

Part b)
P ( Z > 1.2 ) = 1 - P ( Z < 1.2 )
P ( Z > 1.2 ) = 1 - 0.8849
P ( Z > 1.2 ) = 0.1151

Percentage = 0.1151 * 100 = 11.51%

Part c)
P ( Z < -1.2 ) = 0.1151

Percentage = 0.1151 * 100 = 11.51%

Part d)
P ( Z > -1.2 ) = 1 - P ( Z < -1.2 )
P ( Z > -1.2 ) = 1 - 0.1151
P ( Z > -1.2 ) = 0.8849

Percentage = 0.8849 * 100 = 88.49%

Part e)
P ( Z < 0.1 ) = 0.5398

Percentage = 0.5398 * 100 = 53.98%

Part f)
P ( Z > -2.05 ) = 1 - P ( Z < -2.05 )
P ( Z > -2.05 ) = 1 - 0.0202
P ( Z > -2.05 ) = 0.9798

Percentage = 0.9798 * 10 = 97.98%

Part g)
P ( Z > 0.76 ) = 1 - P ( Z < 0.76 )
P ( Z > 0.76 ) = 1 - 0.7764
P ( Z > 0.76 ) = 0.2236

Percentage = 0.2236 * 100 = 22.36%

Part i)
P ( X < 1.5 ) = 0.9332

Percentage = 0.9332 * 100 = 93.32%