Question

The shear strength of each of ten test spot welds is determined, yielding the following data (psi). 364 415 389 367 375 358 409 362 375 365 (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 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 – u/o).] (Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Ginn: Shea tnngt each d t ttApot w 315 365 409 36 389 358 36 335 415 364 ()= Arage n 364t 4IS389 t361 315 358 +409 362+315+365 10 3719 31.9 Psi

Σ G-3), Standad daviatiog n-) n- 1431695 10 ((379) 10 4316951428084. = 20.03 401.2111 PSi 1 J

PCx<x)= 0.5 - 37 १ - 0.95 20 03 314.9 (using alpha tase) = 64s 20.03 33.3935 33.9 412935 411.29 psi (x46)(-1 400-31a9 20.03 P(10) O. 864 3