document.write( "Question 586713: The average number of turkeys per flock in Kennebec County is 38 with a standard deviation of 4.
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document.write( "A. What size flock would have a z score of --1.5?
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document.write( "B. What percent of flocks would have expected to be smaller than A)? \n" );
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Algebra.Com's Answer #373855 by stanbon(75887) ![]() You can put this solution on YOUR website! The average number of turkeys per flock in Kennebec County is 38 with a standard deviation of 4. \n" ); document.write( "A. What size flock would have a z score of --1.5? \n" ); document.write( "Use x = z*s + u \n" ); document.write( "x = -1.5*4 + 38 \n" ); document.write( "x = 32 \n" ); document.write( "----------------------\r \n" ); document.write( "\n" ); document.write( "B. What percent of flocks would have expected to be smaller than A)? \n" ); document.write( "Solution: P(z < -1.5) = normalcdf(-100,-1.5) = 0.0668 = 6.68% \n" ); document.write( "================ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |