Question

In: Statistics and Probability

Suppose that student scores on creativity test are normally distributed. The mean of the test is...

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%

Solutions

Expert Solution

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


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