A manufacturing plant producesstainless steel rods. The lengths of hundreds of r
ID: 2951848 • Letter: A
Question
A manufacturing plant producesstainless steel rods. The lengths of hundreds of rods are measured;the sample mean is 10.508 in and the sample standard deviation is0.011 in. It is assumed that the errors in length are nearlyrandom.
(a)Estimate the probability (%) that a rod taken out of the pile atrandom will have a length less than 10.519 inches.
(b)Estimate the probability (%) that a rod taken out of the pile atrandom will have a length less than 10.492 inches.
(c)The quality assurance engineer specifies that all rods more than±3 from the expected value need to be rejected.Estimate the percentage of rods that will have to be rejected atthis manufacturing plant.
Explanation / Answer
A manufacturing plant produces stainless steelrods. The lengths of hundreds of rods are measured; the samplemean is 10.508 in and the samplestandard deviation is 0.011 in. It isassumed that the errors in length are nearly random.(a)Estimate the probability (%) that a rod taken out of the pile atrandom will have a length less than 10.519 inches.
z = 10.519 - 10.508
0.011 z = +1 (10.519 inches is 1 standarddeviation above themean) P(z < 1)= 0.8413
= 84.13%
(I used a TI-89 calculator, but you can also get it froma z-score table)
(b) Estimate the probability (%) that a rod taken out of thepile at random will have a length less than 10.492inches.
z = 10.492 - 10.508 0.011 z =-1.45 (10.492 inches is 1.45 standarddeviations below themean) P(z <-1.45) = 0.0735 = 7.35% z =-1.45 (10.492 inches is 1.45 standarddeviations below themean) P(z <-1.45) = 0.0735 = 7.35%
(c) The quality assurance engineer specifies that all rods morethan ±3 from the expected value need to berejected.
Estimate the percentage of rods thatwill have to be rejected at this manufacturing plant. You can usethe third part of the Empirical Rule for NormalDistributions: 1. About68% of the data will be within 1 standard deviation from themean (from z = -1 to z = 1) 2. About95% of the data will be within 2 standard deviations from themean (from z = -2 to z =2) 3.About 99.7% of the data willbe within 3 standard deviations from themean (from z = -3 to z = 3) So about0.3% (100%-99.7%) of the rods will be more than 3 standarddeviations from the mean and will need to berejected. So about0.3% (100%-99.7%) of the rods will be more than 3 standarddeviations from the mean and will need to berejected.
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