Question

A specialized machine manufacturer needs to purchase rivets whose strength is greater than 10,000 psi. There are two possible suppliers. Supplier A produces rivets with an average resistance of 14,000 psi with a standard deviation of 2000 psi. Supplier B produces the rivets with an average resistance of 13,000 psi and a standard deviation of 1500 psi. Both populations are normally distributed.

a. Which manufacturer will produce the smallest proportion of “faulty” rivets?

Assume that you decide to buy from manufacturer B:

b. What will be the expected number of defective rivets in a batch of 1000 units?

c. How many rivets on average will have to be tried to find a faulty one?

d. If it is desired that the proportion of defectives does not exceed 1%, where should the average be placed?

Answer 1

ANSWER:

a)

For Supplier A:

µ =    14000      
σ =    2000      
          
P( X ≤    10000   ) = P( (X-µ)/σ ≤ (10000-14000) /2000)  
=P(Z ≤   -2.00   ) =   0.02275

For Supplier B:

µ =    13000      
σ =    1500      
          
P( X ≤    10000   ) = P( (X-µ)/σ ≤ (10000-13000) /1500)  
=P(Z ≤   -2.00   ) =   0.02275

Both are same

b)

Mean = np

=1000*0.0228

=  22.75

c)

Number need to be tried = 1/ 0.02275

= 43.95578902

=46

d)

µ =    13490      
σ =    1500      
          
P( X ≤    10000   ) = P( (X-µ)/σ ≤ (10000-13490) /1500)  
=P(Z ≤   -2.33   ) =   0.00999

Average = 13490

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