8 and are related to 7 A manufacturer of camping gear wishes to buy rivets whose

Question

8 and are related to 7

A manufacturer of camping gear wishes to buy rivets whose breaking strength must exceed 6000 psi. Otherwise, the rivet is considered defective. The breaking strength of rivets is normally distributed. Manufacturer Rosie Riveter (RR) produces rivets with a mean breaking strength of 7500 psi and a standard deviation of 525 psi. Manufacturer Rivet Runs Through It (RT) produces rivets with a mean breaking strength of 6800 psi and a standard deviation of 330 psi. Does RT produce the lowest fraction of defective rivets? Referring to the rivet problem, assume that the camping gear maker buys 75% of its rivets from the manufacturer with the lower fraction of defective rivets, and 25% of its rivets from the other manufacturer, and mixes them together. What is the probability that a rivet in the combined batch will be defective? If you use tables, you will have to interpolate in order to obtain precise probabilities. Referring to the rivet problem, suppose the manufacturer with the higher fraction of defective rivets can improve the process to adjust the standard deviation, but not the mean breaking strength. What new standard deviation should this manufacturer achieve to have the same probability of defective rivets as the other manufacturer (and therefore capture more of the market)?

Explanation / Answer

7) Fraction of defective rivets produced by RR=P(X<=6000)= 0.0021

, using excel function=NORMDIST(6000,7500,525,TRUE)

Fraction of defective rivets produced by RT=P(X<=6000)= 0.0077

, using excel function=NORMDIST(6000,6800,330,TRUE)

So, RT does not produce the lowest fraction of defective rivets.

8) P(bought / RR)=0.75

P(bought / RT)=0.25

P(rivet is defective in combined batch)

= P(bought / RR)*P(defective in RR)+ P(bought / RT)*P(defective in RT)

=0.0021*0.75 + 0.0077*0.25

=0.0035

9) Let be the standard deviation so that Fraction of defective rivets produced by RT=0.0021

Then P(X<=6000)=0.0021

P[(X-6800)/ <= (6000-6800)/] =0.0021

P(Z<z0)=0.0021 where z0=[ (6000-6800)/]

Using excel function =normsinv(0.0021) we have z0=-2.86

So, [ (6000-6800)/]=-2.86

Hence , = -800/(-2.86)=279.72

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