Question

Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for aluminum alloy sheets of a particular type, its mean value and standard deviation are 70 GPa and 1.6 GPa, respectively (values given in the article "Influence of Material Properties Variability on Springback and Thinning in Sheet Stamping Processes: A Stochastic Analysis (Int. J. of Advanced Manuf. Tech, 2010: 117-134)). (a) If X is the sample mean Young's modulus for a random sample of n 16 sheets, where is the sampling distribution of X centered, and what is the standard deviation of the X distribution? EX) - 70 GPa ✓ GPa (b) Answer the questions posed in part (a) for a sample size of n - 256 sheets. EX- GP GPS (c) For which of the two random samples, the one of pat(a) or the one of part (b), is x more likely to be within 1 GPa of 70 GPa? Explain your reasoning. X is more likely to be within 1 GPa of the mean in part (b). This is due to the increased variability of X that comes with a larger sample size X is more likely to be within 1 GPa of the mean in part (a). This is due to the increased variability of X that comes with a smaller sample size X is more likely to be within 1 GPa of the mean in part (a). This is due to the decreased variability of that comes with a smaller sample size X is more likely to be within 1 GPa of the mean in part (b). This is due to the decreased variability of that comes with a larger sample size.

Answer 1

• Soluſion; from the information m the information the wean and Standard deviation of stiffness of aluminium gheets are H=10 Grill of 0 1.6 G Pa (a) For a random sample of m=16 the sanipfing dishobbytien of is [(x ) - Mo = A = to Grey 64 CPA (b) For a random sample of 13056 , Ein) - Mo= M = 70 Gpa =001 Gen

Here, A = o G4 ts = 23 1, PC69느T느키 + = 이쁜 드조스 2회 | 7 - 101 | ( 21이 2 FP-17311 | 1,4391) PC z 1031) - P (2C 12511) - 0.1589 - 0.04lo - 0.171 1 PCs(ced il) 4 n=3, l ) = ( z / 1 - 72 ) = p(2 % 680) I - Plz = 6) . RI), Vecorrib2 = IF (2432-69) b) | (n

P(11?71) = 1-p(7 & 2.60) =1-09999 = 1-0.9953 = 0.0047
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