document.write( "Question 1169956: A Food Marketing Institute found that 30% of households spend more than $125 a week on groceries. Assume the population proportion is 0.3 and a simple random sample of 86 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.37 and 0.49?\r
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document.write( "Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. \n" );
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Algebra.Com's Answer #794806 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! z=(0.37-0.30)/sqrt(0.3*0.7/86) \n" ); document.write( "=0.07/0.0494=1.4166 \n" ); document.write( "z=(0.49-0.30)/0.0494=3.8462 \n" ); document.write( "probability z is between those two values is 0.0782\r \n" ); document.write( "\n" ); document.write( "note, rounding to four decimal places in the above assumes the proportion given will be an integer. It is not, so that the proportion used is a rounding, and four decimal places is not going to be accurate even though it may be calculated. \n" ); document.write( " |