document.write( "Question 248992: Are attitudes toward shopping changing? Sample surveys show that fewer people enjoy shopping than in the past. A recent survey asked a nationwide random sample of 2577 adults if they agreed or disagreed that \"I like buying new clothes, but shopping is often frustrating and time-consuming.\" The population that the poll wants to draw conclusions about is all U.S. residents aged 18 and over. Suppose that in fact 62% of all adult U.S. residents would say \"Agree\" if asked the same questions. What is the probability that 1610 or more of the sample agree?
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Algebra.Com's Answer #181395 by stanbon(75887)\"\" \"About 
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Are attitudes toward shopping changing?
\n" ); document.write( "Sample surveys show that fewer people enjoy shopping than in the past.
\n" ); document.write( "A recent survey asked a nationwide random sample of 2577 adults if they agreed or disagreed that \"I like buying new clothes, but shopping is often frustrating and time-consuming.\"
\n" ); document.write( "The population that the poll wants to draw conclusions about is all U.S. residents aged 18 and over.
\n" ); document.write( "Suppose that in fact 62% of all adult U.S. residents would say \"Agree\" if asked the same questions.
\n" ); document.write( "What is the probability that 1610 or more of the sample agree?
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\n" ); document.write( "population mean = sample mean = 0.62
\n" ); document.write( "sample std = sqrt[0.62*0.38/2577] = 0.0096
\n" ); document.write( "sample proportion = 1610/2577 = 0.6240
\n" ); document.write( "z(0.6240) = (0.6240-0.62)/0.0096 = 0.4167
\n" ); document.write( "P(x >= 1610) = P(z >= 0.4167) = 0.3385
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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