document.write( "Question 289348: Suppose that the wrapper of a candy bar lists its weight as 8 ounces. The actual weights of individual candy bars naturally vary to some extent. Suppose that these actual weights vary according to a normal distribution wth mean 8.5 ounces and standard deviation .325 ounces\r
\n" ); document.write( "\n" ); document.write( "What proportion of he candy bars weigh less less than adverised 8 ounces?\r
\n" ); document.write( "\n" ); document.write( "If the manufacturer wants to decrease this proportion from a) by changing the man of its candy bar weigts, should it increase or decrease that mean?\r
\n" ); document.write( "\n" ); document.write( "If the manufacturer wants to decrease this proportion from a) by changing the standard deviation of its candy bar weight should it increase or decrease that standard deviation?\r
\n" ); document.write( "\n" ); document.write( "Not sure what I need to do ....Thank you for your help
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Algebra.Com's Answer #209594 by stanbon(75887)\"\" \"About 
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Suppose that the wrapper of a candy bar lists its weight as 8 ounces.
\n" ); document.write( "The actual weights of individual candy bars naturally vary to some extent. Suppose that these actual weights vary according to a normal distribution wth mean 8.5 ounces and standard deviation .325 ounces
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\n" ); document.write( "What proportion of the candy bars weigh less less than adverised 8 ounces?
\n" ); document.write( "Find the z-value of 8:
\n" ); document.write( "z(8) = (8-8.5)/0.325 = -1.5385
\n" ); document.write( "P(x < 8) = P(z < -1.5385) = 0.062 weigh less than then 8 ounces.
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\n" ); document.write( "If the manufacturer wants to decrease this proportion from a) by changing the mean of its candy bar weights, should it increase or decrease that mean?
\n" ); document.write( "Increase it.
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\n" ); document.write( "If the manufacturer wants to decrease this proportion from a) by changing the standard deviation of its candy bar weight should it increase or decrease that standard deviation?
\n" ); document.write( "Decrease it to pull the distribution close to the mean.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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