Question

Suppose that student scores on creativity test are normally distributed. The mean of the test is 150 and the standard deviation is 23. Using a z-table (normal curve table), what percentage of students have z-scores a) below 1.63 b) above -0.41 Using a z-table, what scores would be the top and bottom raw score to find the c) middle 50% of students d) middle 10% of students Using a z-table, what is the minimum raw score a student can have on the creativity test and be in the top e) 35% f) 15%

Answer 1

Solution:

Given that,

mean = = 150

standard deviation = = 23

Using standard normal table,

a ) z = 1.63

Using z-score formula,

x = z * +

x = 1.63 * 23 + 150

x = 187.49

The top raw scores = 187.5

b ) z = -0.41

Using z-score formula,

x = z * +

x =  -0.41 * 23 + 150

x = 140.57

The bottom raw scores = 140.6

c ) P(-z < Z < z) = 50%
P(Z < z) - P(Z < z) = 0.50
2P(Z < z) - 1 = 0.50
2P(Z < z ) = 1 + 0.50
2P(Z < z) = 1.50
P(Z < z) = 1.50 / 2
P(Z < z) = 0.75
z = -0.67 and z = 0.67

Using z-score formula,

x = z * +

x = -0.67 * 23 + 150

x = 134.59

x = z * +

x = 0.67 * 23 + 150

x = 165.41

d ) P(-z < Z < z) = 10%
P(Z < z) - P(Z < z) = 0.10
2P(Z < z) - 1 = 0.10
2P(Z < z ) = 1 + 0.10
2P(Z < z) = 1.10
P(Z < z) = 1.10 / 2
P(Z < z) = 0.55
z = -0.13 and z = 0.13

Using z-score formula,

x = z * +

x = -0.13 * 23 + 150

x = 147.01

x = z * +

x = 0.13 * 23 + 150

x = 152.99

e ) P( Z > z) = 35%

P(Z > z) = 0.35

1 - P( Z < z) = 0.35

P(Z < z) = 1 - 0.35

P(Z < z) = 0.65

z = 0.38

Using z-score formula,

x = z * +

x = 0.38 * 23 + 150

x =158.74

f ) P( Z > z) = 15%

P(Z > z) = 0.15

1 - P( Z < z) = 0.15

P(Z < z) = 1 - 0.15

P(Z < z) = 0.85

z =1.04

Using z-score formula,

x = z * +

x = 1.04 * 23 + 150

x =173.92