Question

Instructors assume that test scores follow (approximately) normal distribution, N [68, 10]. Thus if X is a score of a randomly selected student, then X ∼ N [µ = 68, σ = 10] Answer questions below using X-to-Z and Z-to-X conversion rules.

1. Find proportion of students with scores within the interval 50 ≤ X ≤ 75

2. What X-values correspond to z-scores equal to ±1.96

3. Determine the chance that a randomly selected student has score above 60.

4. What proportion of students’ scores would be in the interval (68 ≤ X ≤ 82)?

Answer 1

Solution :

Given that ,

mean = = 68

standard deviation = = 10

a) P(50 x 75)

= P[( 50 - 68 / 10) (x - ) / ( 75 - 68 / 10 ) ]

= P(-1.80 z 0.70)

= P(z 0.70) - P(z -1.80)

Using z table,  

= 0.758 - 0.0359

= 0.7221

b) z ± 1.96

Using z-score formula,

x = z * +

x = -1.96 * 10 + 68

x = 48.4

Using z-score formula,

x = z * +

x = 1.96 * 10 + 68

x = 87.6

c) P(x > 60) = 1 - p( x< 60)

=1- p P[(x - ) / < (60 - 68) / 10 ]

=1- P(z < -0.80)

Using z table,

= 1 - 0.2119

= 0.7881

d) P(68 x 82)

= P[( 68 - 68 / 10) (x - ) / ( 82 - 68 / 10 ) ]

= P( 0 z 1.40)

= P(z 1.40) - P(z 0)

Using z table,  

= 0.9192 - 0.5

= 0.4192