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)     You can put this solution on YOUR website! 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 \n" ); document.write( " |