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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Algebra.Com's Answer #794806 by Boreal(15235)\"\" \"About 
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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.
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