document.write( "Question 1095830: A consumer survey indicates that the average household spends μ =$155 on groceries each week. The distribution of spending amounts is approximately normal with a standard deviation σ =$25. Based on this distribution,\r
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document.write( "What proportion of the population spends more than $175 per week on groceries?\r
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document.write( "How much money do you need to spend on groceries each week to be in the top 20% of the distribution?\r
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document.write( "How much does your family spend per week on groceries, what is your family’s percentile? \n" );
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Algebra.Com's Answer #710336 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z>(x-mean)/sd \n" ); document.write( ">(175-155)/25 or >20/25 \n" ); document.write( "that probability is 0.2119 \n" ); document.write( "------------------- \n" ); document.write( "top 20% is the 80th percentile, and that z is +0.845 \n" ); document.write( "0.845=(x-155)/25 \n" ); document.write( "multiply by 25 both sides \n" ); document.write( "x-155=21.13 \n" ); document.write( "x=$176.13 or more \n" ); document.write( "--------------------- \n" ); document.write( " |