document.write( "Question 586713: 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( "B. What percent of flocks would have expected to be smaller than A)?
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Algebra.Com's Answer #373855 by stanbon(75887)\"\" \"About 
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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
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\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%
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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