Question

Scores on a certain nationwide college entrance examination follow a normal distribution with a mean of 500 and a standard deviation of 100. Find the probability that a student will score...

a) over 650

b) less than 459

c) between 325 and 675

d) If a school only admits students who score over 680, what proportion of the student's pool would be eligible for admission?

e) what limit (score) would you set that makes the top 20% of the students eligible?

Answer 1

Part a)

X ~ N ( µ = 500 , σ = 100 )
P ( X > 650 ) = 1 - P ( X < 650 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 650 - 500 ) / 100
Z = 1.5
P ( ( X - µ ) / σ ) > ( 650 - 500 ) / 100 )
P ( Z > 1.5 )
P ( X > 650 ) = 1 - P ( Z < 1.5 )
P ( X > 650 ) = 1 - 0.9332
P ( X > 650 ) = 0.0668

Part b)

X ~ N ( µ = 500 , σ = 100 )
P ( X < 459 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 459 - 500 ) / 100
Z = -0.41
P ( ( X - µ ) / σ ) < ( 459 - 500 ) / 100 )
P ( X < 459 ) = P ( Z < -0.41 )
P ( X < 459 ) = 0.3409

Part c)

X ~ N ( µ = 500 , σ = 100 )
P ( 325 < X < 675 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 325 - 500 ) / 100
Z = -1.75
Z = ( 675 - 500 ) / 100
Z = 1.75
P ( -1.75 < Z < 1.75 )
P ( 325 < X < 675 ) = P ( Z < 1.75 ) - P ( Z < -1.75 )
P ( 325 < X < 675 ) = 0.9599 - 0.0401
P ( 325 < X < 675 ) = 0.9199

Part d)

X ~ N ( µ = 500 , σ = 100 )
P ( X > 680 ) = 1 - P ( X < 680 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 680 - 500 ) / 100
Z = 1.8
P ( ( X - µ ) / σ ) > ( 680 - 500 ) / 100 )
P ( Z > 1.8 )
P ( X > 680 ) = 1 - P ( Z < 1.8 )
P ( X > 680 ) = 1 - 0.9641
P ( X > 680 ) = 0.0359

Part d)

X ~ N ( µ = 500 , σ = 100 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.2 = 0.8
To find the value of x
Looking for the probability 0.8 in standard normal table to calculate Z score = 0.8416
Z = ( X - µ ) / σ
0.8416 = ( X - 500 ) / 100
X = 584.16
P ( X > 584.16 ) = 0.2