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8the Probabitily What 28g W the theiveople screening process probadiy who positive Type W error? 05j 5r (Round V vour Cosce 1 who perfect havo decimal W that places...

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8the Probabitily What 28g W the theiveople screening process probadiy who positive Type W error? 05j 5r (Round V vour Cosce 1 who perfect havo decimal W that places. "the have W 8 loped V Aabu the 1 discase" spositiveereac avea 1 fairly Check my workProbability 1 the probability 2 Type error? (Round your answeiPrelof 10decimal places.Next

8 the Probabitily What 28g W the theiveople screening process probadiy who positive Type W error? 05j 5r (Round V vour Cosce 1 who perfect havo decimal W that places. "the have W 8 loped V Aabu the 1 discase" spositiveereac avea 1 fairly Check my work Probability 1 the probability 2 Type error? (Round your answei Prel of 10 decimal places. Next



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Measurement error that is normally distributed with a mean of 0 and a standard deviation of 0.5 gram is added to the true weight of a sample. Then the measurement is rounded to the nearest gram. Suppose that the true weight of a sample is 165.5 grams. (a) What is the probability that the rounded result is 167 grams? (b) What is the probability that the rounded result is 167 grams or more?

Why measurement error is normally distributed with me and you equals zero and standard deviation sigma equals 00.5 g. We want to assume that the True Masters 165.5 g and the error is added to the true mass before around it that is whatever we find the measurement error to be. We added to the true mass and then round to the nearest and dearest to the chairman. The way we want to answer a three B based on this information. This question is ultimately testing our understanding of standard normal variables. So to solve, we have to use the information left and right. Information on left is how to map the scores on the probabilities and vice versa for the standard on distribution on the right. Because our random variable is not standard normal rather just simply normal, we must make the conversion between X are variable and Z using Z equals x minus mu over sigma. We have all this information is relevant to solve. So we're seeing a what is probably the rounded results under 67 g. For the resounding results be 1 67. The error must be in either 1 to 2 because with one it gets rounded up to 1 66.5 which is around 167. It must be less than two because at 1 67.5 we rounded up thus Xenon is one minus 0.5 equals two. And Z one is for the probability of the falls between these two scores is from a Z table 20.227 proceeding to be What is the probability that the result is 1 67 or more now, we're simply looking for whether or not the error is greater than one. That probably explains the one is again that probably is is better than two, which is .02-8. Just the tiniest bit larger than part day.

Uh huh. Well that is normally distributed with standard deviation 20 millimeters. The company which is the test the mill hip office, it's not the music was 1 75 consult to and then you greater than 1 75 using and samples We want to calculate the probability of taking a type two error. It truly is 1 85 and A or B or trip. This question is challenging understanding language hypothesis testing in order to determine the probability of making a type two error. For A for all people 20.5 90% were first produced. Critical boundary are critical boundary is the probability is greater than zero equals alpha or 00.5. This is you know, 1.64. That's a critical XZ signal over and piecemeal where mu is the that given our tests, this is 1 85.4 for a critical boundary facing promises test. Thus, data is the probability of Team X greater than 1 85.4. Given that you is 1 85. So this is the probability is less than 25.4 minutes 25 2510 or probably less than 25100.63 equals 0.5 to 51 proceeding to be now for all peoples. 510.5 2016 are actually now one of the 3.2 or 0.161 point 64. The same thus the beta is now the same probability what was one of the 3.2 plugs in? This gives pz less negative 0.28 or 0.3897. Thus in state A is higher than the we can call that increasing annual decrease in beta in the context of this problem.

Uh huh. Well that is normally distributed with standard deviation 20 millimeters. The company which is the test the mill hip office, it's not the music was 1 75 consult to and then you greater than 1 75 using and samples We want to calculate the probability of taking a type two error. It truly is 1 85 and A or B or trip. This question is challenging understanding language hypothesis testing in order to determine the probability of making a type two error. For A for all people 20.5 90% were first produced. Critical boundary are critical boundary is the probability is greater than zero equals alpha or 00.5. This is you know, 1.64. That's a critical XZ signal over and piecemeal where mu is the that given our tests, this is 1 85.4 for a critical boundary facing promises test. Thus, data is the probability of Team X greater than 1 85.4. Given that you is 1 85. So this is the probability is less than 25.4 minutes 25 2510 or probably less than 25100.63 equals 0.5 to 51 proceeding to be now for all peoples. 510.5 2016 are actually now one of the 3.2 or 0.161 point 64. The same thus the beta is now the same probability what was one of the 3.2 plugs in? This gives pz less negative 0.28 or 0.3897. Thus in state A is higher than the we can call that increasing annual decrease in beta in the context of this problem.

The problem here is that the formula is not correct. P of no should be in the denominator instead of P of London. The property, the probably of P of London and then the bar know is 2.33 if you make sure the formula is cracked.


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