document.write( "Question 1161606: The average number of
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document.write( "milligrams (mg) of sodium in a certain brand of low-salt
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document.write( "microwave frozen dinners is 660 mg, and the standard
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document.write( "deviation is 35 mg. Assume the variable is normally
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document.write( "distributed.\r
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document.write( "a. If a single dinner is selected, find the probability that
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document.write( "the sodium content will be more than 670 mg.\r
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document.write( "b. If a sample of 10 dinners is selected, find the probability that the mean of the sample will be larger than
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document.write( "670 mg.\r
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document.write( "c. Why is the probability for part a greater than that for
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document.write( "part b? \n" );
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Algebra.Com's Answer #785197 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z>(x-mean)/sd \n" ); document.write( "=(670-660)/35 \n" ); document.write( "=10/35 or 0.286 \n" ); document.write( "this probability is 0.3875\r \n" ); document.write( "\n" ); document.write( "for 10 dinners, the formula becomes z=(x bar-mean)/sigma/sqrt(n) \n" ); document.write( "and z> 10*sqrt(10)/35=0.90 \n" ); document.write( "that probability is 0.1840\r \n" ); document.write( "\n" ); document.write( " It is much more difficult for the average of 10 things to all be in the same direction compared to a single one.If there is one extreme value, the next probably will be less extreme or even in the other direction. \n" ); document.write( " |