Question

1. Food served at a restaurant should be between 38°C and 49°C when it is delivered to the customer. The process used to keep the food at the correct temperature has a process standard deviation of 2°C and the mean value for these temperature is 42°C.
   1. What is the process capability C_p and C_pk of the process?
   2. Is the process capable? Why or why not?
   3. If the process is not capable, what proportion of the product is nonconforming, assuming a normal distribution of the characteristic?

Answer 1

Answer)

a)

the formula for C_p is (USL - LSL)/(6*standard deviation)

USL = 49 and LSL = 38 and standard deviation = 2

So, C_p = (49-38)/12 = 0.9166666666

And the formula for C_pk is

minimum value of ( (USL - MEAN)/3*STANDARD DEVIATION), (MEAN - LSL)/3*STANDARD DEVIATION))

(USL - MEAN)/3*STANDARD DEVIATION) = (49-40)/(3*2) = 1.5

(MEAN - LSL)/3*STANDARD DEVIATION) = (40-38)/6 = 0.333333

minimum value of (1.5 AND 0.333) is

0.333

so, therefore, C_pk = 0.3

b)

When C_pk is less than 1, then the process is not capable as here C_pk = 0.3 which is less than 1, therefore, given process is not capable

c) As the product is not capable

the proportion of the product which is non conforming can be calculated by multiplying C_pk value with 3

3*0.333 = 0.9 now we need to see the value of z corresponding to 0.9 in -z table which will be our required answer

so proportion = 0.1841