The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 415 362 398 375 364 371 409 389 367 358 (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 psi standard deviation 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.) 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) = ((400 μ)/o).] (Round your answer to four decimal places.)
The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 415 362 398 375 364 371 409 389 367 358 (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 psi standard deviation 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.) 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) = ((400 μ)/o).] (Round your answer to four decimal places.)
MATLAB: An Introduction with Applications
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Author:Amos Gilat
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Please answer all unanswered parts. It’s all part of one question.
![The shear strength of each of ten test spot welds is determined, yielding the following data (psi).
415 362 398 375 364 371
409 389 367 358
(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.)
psi
psi
average
standard deviation
(b) Again assuming a normal distribution, estimate the strength value below which 95% 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.)
psi
(c) Suppose we decide to examine another test spot weld.
obtain the mle of P(X ≤ 400). [Hint: P(X ≤ 400) = ((400
Let X = shear strength of the weld. Use the given data to
μ)/o).] (Round your answer to four decimal places.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3dc90927-a4d4-417e-bb21-8f8d729f068b%2Fb0e9f034-90c5-462e-9ad0-bc3a12e5bbb7%2Fx024ka_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The shear strength of each of ten test spot welds is determined, yielding the following data (psi).
415 362 398 375 364 371
409 389 367 358
(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.)
psi
psi
average
standard deviation
(b) Again assuming a normal distribution, estimate the strength value below which 95% 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.)
psi
(c) Suppose we decide to examine another test spot weld.
obtain the mle of P(X ≤ 400). [Hint: P(X ≤ 400) = ((400
Let X = shear strength of the weld. Use the given data to
μ)/o).] (Round your answer to four decimal places.)
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