The article \"Characterization of Room TemperatureDamping in Aluminu-Indium Allo

Question

The article "Characterization of Room TemperatureDamping in Aluminu-Indium Alloys" suggests that x = A1 matrix grain size (m) for an alloy consistingof 2% indium could be modeled with a normal distribution having = 96 and = 14. What proportion ofgrain sizes are at least 75? The article "Characterization of Room TemperatureDamping in Aluminu-Indium Alloys" suggests that x = A1 matrix grain size (m) for an alloy consistingof 2% indium could be modeled with a normal distribution having = 96 and = 14. What proportion ofgrain sizes are at least 75?

Explanation / Answer

Just focus on the = 96 and = 14. Since it's a normal distribution, you can either use yourcalculator or a standard table of normal probabilities. Calculator Method (for TI-83 and TI-84): 2nd Vars (or blue lettered DISTR) navigate to: "normalcdf(" normalcdf(lower bound, upper bound, mean, standarddeviation) so.. normalcdf(75, 9999999, 96, 14)   <-- withoutspaces on the calculator (the upper bound is just a bunch of 9's torepresent infinitiy) it would give you a Proportion of .9332 or 93.32% If you use a standard table of normal probabilities: (x - ) / = z-score in this case: (75-96) / 14 = -1.5 using the table, it would give you a proportion of .0668 or6.68% However, that's the proportion of grains less than 75. So you would take 1 and subtract the proportion you've foundfrom it. 1 - .0668 = .9332 or 93.32% That gives you the proportion of grains that are at least 75in size normalcdf(lower bound, upper bound, mean, standarddeviation) so.. normalcdf(75, 9999999, 96, 14)   <-- withoutspaces on the calculator (the upper bound is just a bunch of 9's torepresent infinitiy) it would give you a Proportion of .9332 or 93.32% If you use a standard table of normal probabilities: (x - ) / = z-score in this case: (75-96) / 14 = -1.5 using the table, it would give you a proportion of .0668 or6.68% However, that's the proportion of grains less than 75. So you would take 1 and subtract the proportion you've foundfrom it. 1 - .0668 = .9332 or 93.32% That gives you the proportion of grains that are at least 75in size

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