Question

A manufacturer produces a line of tableware made from sand casting a special alloy of several metals. After casting, the pieces go through a series of shaping, grinding, buffing, and polishing steps. The manufacturer recently began a program to study the total time spent on the grinding, buffing, and polishing steps across their entire product line. The distribution surrounding the random variable representing the total time (in minutes) spent in grinding, buffing, and polishing can be modeled by a normal distribution with6.48=μ and variance=6892.

a.Management has given the production supervisor a goal of having parts complete the grinding, buffing, and polishing operations in less than 1 hour. What is the probability of a part completing all three operations in less than 1 hour? What can the production supervisor tell management about their current status in meeting the goal? (2 points)

b.The production supervisor would like to know the number of minutes for which 90% of the parts have a grinding, buffing, and polishing time less than or equal to this value. He needs this information so that he can report to management that “90% of our parts complete grinding, buffering, and polishing in no more than X minutes.” What should he report to management? (2 points)

Answer 1

Let X be the total time in minutes        
X follows normal distribution mean μ and standard deviation σ        
Given μ = 64.8  σ = sqrt(variance) = sqrt(6892) = 83.0181     
         
a) To find P(a part completing all three operations in less than 1 hour)        
that is to find P(X < 60 minutes)        
We use Excel function NORM.DIST to find the probability        
P(X < 60) = NORM.DIST(60, 64.8, 83.0181, TRUE)        
                   = 0.4769        
P(a part completing all three operations in less than 1 hour) = 0.4769        
         
b) Let X' be the number of minutes for which 90% of the parts have a total time less than or equal to this value        
Then P(X < X') = 0.90        
We use Excel function NORM.INV to find X'        
X' = NORM.INV(0.9, 64.8, 83.0181)        
                   = 171.192        
No. of minutes for which 90% of the parts have a total time less than or equal to this value = 171.19 minutes