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# 5 pts 136 p42 suppose the flow of current milliamps

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Unformatted text preview: have a value of 0.9990, any of the values are correct (1.5 pts.) 1.36 (p.42). suppose the flow of current (milliamps) in wire strips of a certain type under specified conditions can be modeled with a normal distribution having μ = 20 and = 1. a) What proportion of strips will have a current flow of between 18.5 and 22 milliamps? proportion (18.5 &lt; x &lt; 22) = proportion (-1.5 &lt; z &lt; 2) = proportion (z &lt; 2) – proportion (z &lt; -1.5) = 0.9772 – 0.0668 = 0.9104 6 b) What proportion of strips will have a current flow exceeding 15 milliamps? The table only goes down to -3.89, the proportion to the left of -3.89 = 0.0000. Since prop (z &lt; -5) &lt; prop (z &lt; -3.89), prop (z &lt; -5) = 0.0000. prop (z &gt; -5) = 1 – prop (z &lt; -5) = 1 – 0 = 1. c) How large must a current flow be to be among the largest 5% of all flows? proportion (z &lt; c) = 0.95 ==&gt; c = 1.645 (or 1.64 or 1.65) (2 pts.) 1.38 (p.42). An article suggest that x = Al (Aluminum) matrix grain size (...
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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 University-West Lafayette.

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