Question

Izmir Dokuz Eylul university students ' height is 168cm average and 10cm
normal with standard deviation. According to this, the height of a randomly selected student:
a) be 165cm (5 points)
B) 165 cm long to be long (5 points)
c) 165 cm short (5 points)
d) to be between 150cm and 160cm (10 points)
Calculate their probability.

Answer 1

Part a)

X ~ N ( µ = 168 , σ = 10 )

P ( X = 165 ) =   P ( 164.5 < X < 165.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 164.5 - 168 ) / 10
Z = -0.35
Z = ( 165.5 - 168 ) / 10
Z = -0.25
P ( -0.35 < Z < -0.25 )
P ( 164.5 < X < 165.5 ) = P ( Z < -0.25 ) - P ( Z < -0.35 )
P ( 164.5 < X < 165.5 ) = 0.4013 - 0.3632
P ( 164.5 < X < 165.5 ) = 0.0381

Part b)

X ~ N ( µ = 168 , σ = 10 )
P ( X > 165 ) = 1 - P ( X < 165 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 165 - 168 ) / 10
Z = -0.3
P ( ( X - µ ) / σ ) > ( 165 - 168 ) / 10 )
P ( Z > -0.3 )
P ( X > 165 ) = 1 - P ( Z < -0.3 )
P ( X > 165 ) = 1 - 0.3821
P ( X > 165 ) = 0.6179

Part c)

X ~ N ( µ = 168 , σ = 10 )
P ( X < 165 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 165 - 168 ) / 10
Z = -0.3
P ( ( X - µ ) / σ ) < ( 165 - 168 ) / 10 )
P ( X < 165 ) = P ( Z < -0.3 )
P ( X < 165 ) = 0.3821

Part d)

X ~ N ( µ = 168 , σ = 10 )
P ( 150 < X < 160 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 150 - 168 ) / 10
Z = -1.8
Z = ( 160 - 168 ) / 10
Z = -0.8
P ( -1.8 < Z < -0.8 )
P ( 150 < X < 160 ) = P ( Z < -0.8 ) - P ( Z < -1.8 )
P ( 150 < X < 160 ) = 0.2119 - 0.0359
P ( 150 < X < 160 ) = 0.1759