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A what proportion of grain sizes exceed 100

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Unformatted text preview: μm) for an alloy consisting of 2% indium could be modeled with a normal distribution having μ = 96 and = 14. a) What proportion of grain sizes exceed 100? proportion (96 < x) = proportion (0.29 < z) =1 - proportion (z < 0.29) = 1 – 0.6141 = 0.3859 b) What proportion of grain sizes are at least 75? proportion (75 < x) = proportion (-1.5 < z) =1 - proportion (z < -1.5) = 1 – 0.0668 = 0.9332 c) What proportion of grain sizes are between 50 and 75? proportion (50 < x < 75) = proportion (-3.29 < z < -1.5) = proportion (z < -1.5) – proportion (z < 3.29) = 0.0668 – 0.0005 = 0.0663 d) What interval (a,b) includes the central 90% of all grain sizes (so that 5% are below a and 5% are above b)? As we discussed in class, we are looking for the 0.90 + 0.05 = 0.95 percentile. From 1.36c, z = 1.645 or 1.64 or 1.65. Since the standard normal cu...
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This note was uploaded on 01/29/2014 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue.

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