I am having some trouble with this homework question. What can I have done right? The shear strength of each of ten test spot welds is determined, yielding the following data (psi).402362367406409389358375415365(a) Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood. (Round your answers to two decimal places.)average384.8psistandard deviation21.87X psi(b) Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. [Hint: What is the 95th percentile in terms of u and o? Now use the invariance principle.] (Round your answer to two decimal places.)420.78X psi(c) Suppose we decide to examine another test spot weld. Let X = shear strength of the weld. Use the given data to obtain the mle of P(X < 400). [Hint: P(X < 400) = Þ((400 – µ)/o).] (Round your answer to four decimal places.)0.7565

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The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 402 362 367 406 389 358 375 415 365 (a) Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood. (Round your answers to two decimal places.) average 384.8 V psi standard deviation 21.87 X psi (b) Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. [Hint: What is the 95th percentile in terms of u and o? Now use the invariance principle.] (Round your answer to two decimal places.) 420.78 X psi (c) Suppose we decide examine another test spot weld. Let X- shear strength of the weld. Use the given data to obtain the mle of P(X s 400). [Hint: P(X s 400) - 0((400 - u)/a).] (Round your answer to four decimal places.) 0.7565

The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 402 362 367 406 389 358 375 415 365 (a) Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood. (Round your answers to two decimal places.) average 384.8 V psi standard deviation 21.87 X psi (b) Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. [Hint: What is the 95th percentile in terms of u and o? Now use the invariance principle.] (Round your answer to two decimal places.) 420.78 X psi (c) Suppose we decide examine another test spot weld. Let X- shear strength of the weld. Use the given data to obtain the mle of P(X s 400). [Hint: P(X s 400) - 0((400 - u)/a).] (Round your answer to four decimal places.) 0.7565

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Question

I am having some trouble with this homework question. What can I have done right?

The shear strength of each of ten test spot welds is determined, yielding the following data (psi).
402
362
367
406
409
389
358
375
415
365
(a) Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood. (Round your answers to two decimal places.)
average
384.8
psi
standard deviation
21.87
X psi
(b) Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. [Hint: What is the 95th percentile in terms of u and o? Now use the invariance principle.] (Round your answer to two decimal places.)
420.78
X psi
(c) Suppose we decide to examine another test spot weld. Let X = shear strength of the weld. Use the given data to obtain the mle of P(X < 400). [Hint: P(X < 400) = Þ((400 – µ)/o).] (Round your answer to four decimal places.)
0.7565

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