The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 367 362 415 358 367 373 409 387 389 375 (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 standard deviation psi 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)/o).] (Round your answer to for decimal places.)

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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Chapter10: Statistics
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The shear strength of each of ten test spot welds is determined, yielding the following data (psi).
367 362 415 358 367 373
409 387 389
375
(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
standard deviation
psi
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) = Þ((400 – µ)/o).] (Round your answer to four
decimal places.)
Transcribed Image Text:The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 367 362 415 358 367 373 409 387 389 375 (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 standard deviation psi 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) = Þ((400 – µ)/o).] (Round your answer to four decimal places.)
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