Question

Part B

Q5. [6] 64% of Ball State University students are inter-state students. If we observe 100 students at random, use the normal approximation to find the probability of observing (a) more than 70 inter-state students. (b) not more than 50 inter-state students.

Answer 1

Mean = n * P = ( 100 * 0.64 ) = 64
Variance = n * P * Q = ( 100 * 0.64 * 0.36 ) = 23.04
Standard deviation = $\sqrt{variance} = \sqrt{23.04}$ = 4.8

Part a)

P ( X > 70 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 70 + 0.5 ) = P ( X > 70.5 )

$X \sim N ( \mu = 64 , \sigma = 4.8 )$
P ( X > 70.5 ) = 1 - P ( X < 70.5 )
Standardizing the value
$Z = ( X - \mu ) / \sigma$
Z = ( 70.5 - 64 ) / 4.8
Z = 1.35
$P ( ( X - \mu ) / \sigma ) > ( 70.5 - 64 ) / 4.8 )$
P ( Z > 1.35 )
P ( X > 70.5 ) = 1 - P ( Z < 1.35 )
P ( X > 70.5 ) = 1 - 0.9115
P ( X > 70.5 ) = 0.0885

Part b)

P ( X <= 50 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 50 + 0.5 ) = P ( X < 50.5 )

$X \sim N ( \mu = 64 , \sigma = 4.8 )$
P ( X < 50.5 )
Standardizing the value
$Z = ( X - \mu ) / \sigma$
Z = ( 50.5 - 64 ) / 4.8
Z = -2.81
$P ( ( X - \mu ) / \sigma ) < ( 50.5 - 64 ) / 4.8 )$
P ( X < 50.5 ) = P ( Z < -2.81 )
P ( X < 50.5 ) = 0.0025