document.write( "Question 1183214: The mean annual cost of automobile insurance is R95. Assume that the population standard deviation is R14. What is the probability that a simple random sample of size 30 for automobile insurance policies will have a sample mean less than R90 \n" ); document.write( "
Algebra.Com's Answer #813474 by Boreal(15235)\"\" \"About 
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this is a z-test with z=(x bar-mean)/sigma/sqrt (n)
\n" ); document.write( "z < (90-95)/14/sqrt(30)
\n" ); document.write( "z < -5*sqrt(30)/14 inverting and multiplying to deal with division.
\n" ); document.write( "z < -1.956
\n" ); document.write( "0.0252 is probability
\n" ); document.write( "The sd of the sampling distribution is 14/sqrt(30) or 2.556
\n" ); document.write( "Then it is z < -5/2.556
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