document.write( "Question 1074944: Suppose a simple random sample of size n = 36 is taken from a population with mean = 64 and standard deviation = 18.\r
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document.write( "a) What are the mean and standard deviation of the sampling distribution?
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document.write( "b) What is P(x < 62.6)?\r
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document.write( "I generally can figure this out if given the formulas but I don't know the formulas in this instance. \n" );
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Algebra.Com's Answer #689635 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! The mean of a sampling distribution is the mean of the population from which it was taken. mean=64 \n" ); document.write( "the sd of a sampling distribution is the sd of the population divided by the square root (n) \n" ); document.write( "This would be 18/sqrt(36)=3. The sampling distribution is N~n=36 (64,3) \n" ); document.write( "The probability p<62.6 is a z \n" ); document.write( "z=(x bar-mean)/s/sqrt (n) \n" ); document.write( "z<(62.6-64)/3 \n" ); document.write( "z<-1.4/3 \n" ); document.write( "z< -0.467. \n" ); document.write( "p=0.3202 \n" ); document.write( " |