Question

An automatic machine in a manufacturing process is operating properly if the lengths of an important component are normally distributed with a mean of 110 cm and a standard deviation of 4.8 cm.

a) Find the probability that a selected component is longer than 112 cm.

b) Find the probability that if 3 are randomly selected, all have lengths less than 112 cm.

c) Which length marks the longest 15% of the subcomponents?

Answer 1

X be the length of the components      
X follows normal distribution mean μ and standard deviation σ      
Given μ = 110 cm  σ = 4.8   
       
a) To find P(component is longer than 112 cm)      
that is to find P(X > 112)      
P(X > 112) = 1 - P(X ≤ 112)      
We use Excel function NORM.DIST to find the probability      
                     = 1 - NORM.DIST(112, 110, 4.8, TRUE)      
                    = 1 - 0.6615      
                    = 0.3385      
P(component is longer than 112 cm) = 0.3385      
       
b) First we find P(one component is less than 112 cm)      
that is to find P(X < 112)      
We use Excel function NORM.DIST to find the probability      
P(X < 112) = NORM.DIST(112, 110, 4.8, TRUE)      
                    = 0.6615      
P(3 components selected have lengths less than 112 cm)      
   = 0.6615 * 0.6615 * 0.6615     
   = 0.2895     
P(3 components selected have lengths less than 112 cm) = 0.2895      
       
c) Let X' be the length which marks the longest 15% of the subcomponents.      
Then P(X > X') = 0.15      
that is P(X ≤ X') = 1 - 0.15 = 0.85      
We use Excel function NORM.INV to find X'      
X' = NORM.INV(0.85, 110, 4.8, TRUE)      
X' = 114.9749 cm      
The length that marks the longest 15% of the sub-components = 114.9749 cm.