document.write( "Question 1140729: Consider a normal population with a mean of 50 and standard deviation of 2. A random sample of size 4 is selected from the population. Find the probability that the sample mean is between 49.1 and 50.15. \n" ); document.write( "
Algebra.Com's Answer #761448 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z=(x bar-mean)/sigma/sqrt(n) \n" ); document.write( "so z1=-0.9/2/sqrt(4)=-0.9*2/2=-0.9 (invert the denominator and multiply, although it doesn't matter here) \n" ); document.write( "z2=0.15*2/2=0.15 \n" ); document.write( "That probability, z between -0.9 and +0.15, is 0.3756 \n" ); document.write( " |