document.write( "Question 1161601: The average individual monthly spending in
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document.write( "the United States for paging and messaging services
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document.write( "is $10.15. If the standard deviation is $2.45 and the
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document.write( "amounts are normally distributed, what is the
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document.write( "probability that a randomly selected user of these
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document.write( "services pays more than $15.00 per month? Between
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document.write( "$12.00 and $14.00 per month? \n" );
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Algebra.Com's Answer #787510 by Boreal(15235) You can put this solution on YOUR website! z=(x-mean)/sd \n" ); document.write( "z=(15-10.15)/2.45=1.978 \n" ); document.write( "probability z> 1.978 is 0.0239 \n" ); document.write( "between 12 and 14 \n" ); document.write( "do it the same way or on calculator 2nd VARS2 normal cdf ENTER (12,14,10.15,2.45)ENTER \n" ); document.write( "0.1671 \n" ); document.write( "it is probability z is between 0.755 and 1.571 (0.1670), the difference in the last decimal place was that I was rounding in the second. \n" ); document.write( " |