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Factory - produces plate glass with mean thickness of Amm and standard Type numbers in the boxes deviation of Llmm: A simple random sample of 1OO sheets of glass is...

Question

Factory - produces plate glass with mean thickness of Amm and standard Type numbers in the boxes deviation of Llmm: A simple random sample of 1OO sheets of glass is to be 10 points measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 3.91 mm?Round your answers to 5 decimal places;

factory - produces plate glass with mean thickness of Amm and standard Type numbers in the boxes deviation of Llmm: A simple random sample of 1OO sheets of glass is to be 10 points measured, and the mean thickness of the 100 sheets is to be computed. What is the probability that the average thickness of the 100 sheets is less than 3.91 mm? Round your answers to 5 decimal places;



Answers

The fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation 2. a. What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds $14.5 ?$ b. Find an interval that includes, with probability $0.95,$ the average fracture strength of 100 randomly selected pieces of this glass.

This question involves some central limit theorem in some empirical rule. So when we take our sample of 100 the distribution should be normal with a center around the population mean I've used um you here for that, and then the standard deviation of our samples is going to be the population standard deviation, which it says is 10 pcs Divided by the square root of our sample size, which is 100 And 10, divided by the square root of 100 is one. So when I come here and number my number, line by one, It's going to be my population average plus one to the right, population average -1 to my left. And the empirical rule says that when you are within one standard deviation of the mean below or above 68% Of the result should be within that range. So 68% is our answer.

Now here on this problem, we are told that we have bottles coming off the production line. There's a 10 chance that one of them as serious false. What that tells us that P has 0.10 has changed percentage. So that's what we are also told. We are picking two models randomly. And so that means in as we wrote to now, this is a binomial distribution which is anytime you pick one, there are only two options. It is either flawed or not flawed. So since there's only a binary chances in the binomial distribution and the expected value for binomial is in times people, It's going to calculate that. That would be 0.10 times two What does point to zero. And so the expected value at this .20, the variance for binomial is in times P Sometimes one -P. Me too, I was .10, I was .90. So again, we just calculate this. We just talked that's freedom. Yeah, Then gives a 0.18. That's our variance is 0.1

So, so in this question we are asked to determine the mean and variance of lengths of a discreet Uniform distribution starting at 5 19.0, going all the way to 5 90.9. So meaning is just the final value plus the middle value divided by two. So 5 90 .9 Plus 5 90 Over two is basically 5 90.45. That's what I mean. And our standard tv or experiences being minus mm plus one squared minus 1/12. So 5 19.9 -5 90 is .9 0.9 Plus one is 1.9. 1.9 Squared is 3.61 minus one is basically 2.61 over 12, Which gives us the variance of .2175.

Draw normal, Kurt. The meanness 100 in the standard deviation is 10. You will be Addington each standard Eve. You shouldn't go up, which three of them, and subtract in each tentative you should going down now. The problem is asking by the probability that a selected value at random is at least 100 meaning we want something 100 or bigger. And if you notice this, cuts are normal. Distribution exactly in half and half is often represented by the percentage lengthy person.


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