Question
A company that produces rivets used by commercial aircraft manufactures knows that the shearing strength (force required to breaking) of its rivets is a major concern. The company believes the shearing strength of its rivets is normally distributed with a mean of 925 lbs and a standard deviation of 18 lbs.
A. If the company is correct, what percentage of its rivets have a shearing strength greater than 900 lbs?
B. What is the upper bound for the shearing strength of the weakest 1% of its rivets?
C. If one rivet is randomly selected from all of the rivets, what is the probability that it will require a force of at least 920 lbs to break it?
A. If the company is correct, what percentage of its rivets have a shearing strength greater than 900 lbs?
B. What is the upper bound for the shearing strength of the weakest 1% of its rivets?
C. If one rivet is randomly selected from all of the rivets, what is the probability that it will require a force of at least 920 lbs to break it?
Answers
A. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score. Multiply by 100.
B. Use same table to find Z for P = .01. Insert in above equation and calculate.
C. Process same as A, but don't multiply by 100.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score. Multiply by 100.
B. Use same table to find Z for P = .01. Insert in above equation and calculate.
C. Process same as A, but don't multiply by 100.