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Suppose that the fracture angle under pure compression of a randomly selected specimen of fiberreinforced polymer-matrix composite material is normally distributed with mean value 53 andstandard deviation 1 (suggested in the article "Stochastic Failure Modelling of Un directional)Composite Ply Failure," Reliability Engr. and System Safety, $1-9 ;$ this type of material iComposite Ply Failure," Reliability Engr. and System Safety, $2012 : 1-9 ;$ this type of material is used extensively in the aerospace industry).(a) If a random sample of 4 specimens is selected, what is the probability that the sample mean fracture angle is at most 54 ? Between 53 and 54 ?(b) How many such specimens would be required to ensure that the first probability in (a) is a least. 999$?$

(a) .9772, .4772 (b) 10

Intro Stats / AP Statistics

Chapter 4

Joint Probability Distributions and Their Applications

Section 1

Jointly Distributed Random Variables

Probability Topics

The Normal Distribution

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told that the fracture angle under pure compression of a randomly selected specimen of fiber reinforced polymer matrix composite material, is normally distributed with mean value three and a standard deviation of one in part A. We're told that a random sample of four specimens is selected and were asked to find the probability that thes sample mean fracture angle is at most 54. Well, well. Let ex partner with sample mean fracture angle. Yeah, of the four samples Now, the individual fracture angles are normally distributed, so it follows that X bar is also a normal random variable, in particular with a mean the expected value of X bar is the same as the mean mu which were given is 53 but with the standard deviation standard deviation of X bar. Well, this is going to be the standard deviation of the population Sigma over a squared of the number of samples and were given that sigma is one. And the end was what we had for samples root for this is one half or 0.5. Therefore, it follows that ability at this is less than or equal to 54. Well, this is the same as the probability At the standard normal vector real, very random variable Z is less than or equal to 54 minus the mean 53 over the standard deviation 0.5, which is the same as probability that Z is less than or equal to two, which is the same as 52 And using a table or computer, we calculate that this is 0.977 to approximately were also asked to find probability that the sample mean fracture angle lies between 53 and 54. So the probability that X bar lies between 53 54. Well, this is the probability at the standard normal random variables. E lies between 53 minus the mean 53 over the steering deviation five and 54 minus the mean 53 over the sneering deviation 530.5, which is the same as the probability that Z lies between zero and to which, of course, is the same as five to minus 50 And we can either calculate thes using computer or looked in that book. We find that these air approximately 24 772 This could also be found recognizing the 50 is 500.5 and subtracting this from our previous result, 0.9772 Next, in part B, grass defined how many specimens will be required to ensure that the first probability, in part a probability that the sample mean fracture angle is at most 54 is at least 0.999 So to do this will simply of police in our formulas from part a four with n and most of the probability expression equal 2.999 So we have 0.999 We want this to be equal to at least approximately probability that X bar is less than or equal to 54 and we have that this is equal to probability. At the standard, normal variable Z is less than or equal to 54 minus the mean, which stays the same. Still 53 over the standard deviation, which is now stare deviation of the distribution, which is one over a squared of the number of specimens. And and this is, of course, the same as five of approximately or five en, and therefore again using a table or computer, we find that Route N is approximately 3.9 and therefore squaring end. We find that N is approximately 9.5. Now we have that warm taking a sample. We have two useful numbers, so it has to be a whole number. Let's round up. So you want to take X to be the ceiling of N, which is 10 so

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