document.write( "Question 427216: The weight of food packed in certain containers has a mean of 16 ounces and a standard deviation of 0.6 ounces. There are 36 containers placed in a box for shipping. Find the probability that a randomly picked box will weight over 580 ounces. \n" ); document.write( "
Algebra.Com's Answer #297132 by robertb(5830)\"\" \"About 
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Let X = r.v. representing the weight of certain containers.
\n" ); document.write( "Then the distribution in question is Y = 36X. This is also a normal distribution with mean E(Y) = E(36X) = 36E(X) = 36*16 = 576, and variance Var(Y) = Var(36X) = \"36%5E2Var%28X%29+=+1296Var%28X%29+=+1296%280.6%29%5E2\" = 466.56. The standard deviation is thus \"sqrt%28466.56%29+=+21.6\". You're looking for the probability \"P%28Y+%3E+580%29+=+P%28Z+=+%28Y+-+576%29%2F21.6+%3E+%28580+-+576%29%2F21.6+=+0.185%29\", or P(Z > 0.185). Now use any table of standard normal probabilities.
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