Question

3. The mean and standard deviation price of a Laptop is $4500 and $ 225. Find the probability that the price of Laptop is :

a) Between $4000 and $4750

b) More than $5000

Answer 1

Part a)
X ~ N ( µ = 4500 , σ = 225 )
P ( 4000 < X < 4750 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 4000 - 4500 ) / 225
Z = -2.2222
Z = ( 4750 - 4500 ) / 225
Z = 1.1111
P ( -2.22 < Z < 1.11 )
P ( 4000 < X < 4750 ) = P ( Z < 1.11 ) - P ( Z < -2.22 )
P ( 4000 < X < 4750 ) = 0.8667 - 0.0131
P ( 4000 < X < 4750 ) = 0.8536

Part b)
X ~ N ( µ = 4500 , σ = 225 )
P ( X > 5000 ) = 1 - P ( X < 5000 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 5000 - 4500 ) / 225
Z = 2.2222
P ( ( X - µ ) / σ ) > ( 5000 - 4500 ) / 225 )
P ( Z > 2.2222 )
P ( X > 5000 ) = 1 - P ( Z < 2.2222 )
P ( X > 5000 ) = 1 - 0.9869
P ( X > 5000 ) = 0.0131