document.write( "Question 1148747: In a certain small town, Comcast has 100000 subscribers. According to Census data, 40 percent of the subscribers are single parent households. A simple random sample of 400 households were selected for a survey. The percentage of single parent households in the sample will be around
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\n" ); document.write( "\n" ); document.write( "Estimate the chance that between 35 percent and 45 percent of the households in the sample are single parent households. ___
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Algebra.Com's Answer #770124 by rothauserc(4718)\"\" \"About 
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standard error of the sample = square root(0.40 * (1-0.40)/400) = 0.0245
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\n" ); document.write( "The percentage of single parent households in the sample will be around
\n" ); document.write( "40 give or take 0.0245 or so.
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\n" ); document.write( "Since the sample size is greater than 30, we can use the normal distribution's z-tables
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\n" ); document.write( "Probability (P) (0.35 < X < 0.45) = P(X < 0.45) - P(X < 0.35)
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\n" ); document.write( "z-score(0.45) = (0.45 - 0.40)/0.0245 = 2.04
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\n" ); document.write( "P(X < 0.45) = 0.9793
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\n" ); document.write( "z-score(0.35) = (0.35 - 0.40)/0.0245 = -2.04
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\n" ); document.write( "P(X < 0.35) = 0.0207
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\n" ); document.write( "P(0.35 < X < 0.45) = 0.9793 - 0.0207 = 0.9586 is approximately 0.96 or 96%
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