Question

Consider the population of all 1-gallon cans of dusty rose paint manufactured by a particular paint company. Suppose that a normal distribution with mean μ = 4 ml and standard deviation σ = 0.2 ml is a reasonable model for the distribution of the variable x = amount of red dye in the paint mixture. Use the normal distribution model to calculate the following probabilities. (Round your answers to four decimal places.)

(a) P(x < 4) =  

(b) P(x < 4.2)=  

(c) P(x ≤ 4.2) =

(d) P(3.6 < x < 4.2) =  

(e) P(x > 3.5) =  

(f) P(x > 3) =

*PLEASE SHOW WORK. THANK YOU!

Answer 1

a)

X ~ N ( µ = 4 , σ = 0.2 )  
P ( X < 4 ) = ?  
Standardizing the value   
Z = ( X - µ ) / σ  
Z = ( 4 - 4 ) / 0.2  
Z = 0  
P ( ( X - µ ) / σ ) < ( 4 - 4 ) / 0.2 )  
P ( X < 4 ) = P ( Z < 0 )   
P ( X < 4 ) = 0.5

b)

X ~ N ( µ = 4 , σ = 0.2 )
P ( X < 4.2 ) = ?
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 4.2 - 4 ) / 0.2
Z = 1
P ( ( X - µ ) / σ ) < ( 4.2 - 4 ) / 0.2 )
P ( X < 4.2 ) = P ( Z < 1 )
P ( X < 4.2 ) = 0.8413

c)

P( X <= 4.2) = ?

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 4.2 - 4 ) / 0.2
Z = 1
P ( ( X - µ ) / σ ) < ( 4.2 - 4 ) / 0.2 )
P ( X < 4.2 ) = P ( Z < 1 )
P ( X < 4.2 ) = 0.8413

d)

P ( 3.6 < X < 4.2 ) = ?
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 3.6 - 4 ) / 0.2
Z = -2
Z = ( 4.2 - 4 ) / 0.2
Z = 1
P ( -2 < Z < 1 )
P ( 3.6 < X < 4.2 ) = P ( Z < 1 ) - P ( Z < -2 )
P ( 3.6 < X < 4.2 ) = 0.8413 - 0.0228
P ( 3.6 < X < 4.2 ) = 0.8186

e)

P ( X > 3.5 ) = 1 - P ( X < 3.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 3.5 - 4 ) / 0.2
Z = -2.5
P ( ( X - µ ) / σ ) > ( 3.5 - 4 ) / 0.2 )
P ( Z > -2.5 )
P ( X > 3.5 ) = 1 - P ( Z < -2.5 )
P ( X > 3.5 ) = 1 - 0.0062
P ( X > 3.5 ) = 0.9938

f)

P ( X > 3 ) = 1 - P ( X < 3 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 3 - 4 ) / 0.2
Z = -5
P ( ( X - µ ) / σ ) > ( 3 - 4 ) / 0.2 )
P ( Z > -5 )
P ( X > 3 ) = 1 - P ( Z < -5 )
P ( X > 3 ) = 1 - 0
P ( X > 3 ) = 1