document.write( "Question 456318: QUESTIONS 1-5 ARE BASED ON THE FOLLOWING INFORMATION
\n" ); document.write( "The fill in bottles of a wine is normally distributed with mean μ = 750 milliliters (ml) and standard deviation of σ = 9 ml.
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\n" ); document.write( "1 For samples of size n = 64 bottles selected from this population, the mean of the sampling distribution of x̄ is _____ and the standard error of x̄ ______.
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\n" ); document.write( "a E(x̄) = 93.75 se(x̄) = 9
\n" ); document.write( "b E(x̄) = 750 se(x̄) = 9
\n" ); document.write( "c E(x̄) = 750 se(x̄) = 0.141
\n" ); document.write( "d E(x̄) = 750 se(x̄) = 1.125
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\n" ); document.write( "2 \"The probability that the mean of a sample of n = 64 bottles falls between 748 and 752 ml is,
\n" ); document.write( "P(748 < x̄ < 752) = _____.\"
\n" ); document.write( "
\n" ); document.write( "a 0.9500
\n" ); document.write( "b 0.9250
\n" ); document.write( "c 0.8165
\n" ); document.write( "d 0.7154
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\n" ); document.write( "3 The proportion of the means obtained from samples of size n = 64 bottles that fall within ±1 ml from the mean of the population of bottles is _____.
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\n" ); document.write( "a 0.6827
\n" ); document.write( "b 0.6266
\n" ); document.write( "c 0.5223
\n" ); document.write( "d 0.4039
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\n" ); document.write( "4 The margin of error for the interval which contains 80% of all sample means is: MOE = _______.
\n" ); document.write( "a 2.21 ml.
\n" ); document.write( "b 1.85 ml.
\n" ); document.write( "c 1.44 ml.
\n" ); document.write( "d 1.29 ml.
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\n" ); document.write( "5 To obtain a margin of error MOE = ± 1 ml for an interval that contains 95% of all sample means, the sample size is n = _____.
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\n" ); document.write( "a 312
\n" ); document.write( "b 217
\n" ); document.write( "c 159
\n" ); document.write( "d 122
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Algebra.Com's Answer #313234 by edjones(8007)\"\" \"About 
You can put this solution on YOUR website!
1)s=9/sqrt(64)=9/8=1.125
\n" ); document.write( "m=750
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\n" ); document.write( "2)
\n" ); document.write( "z=2/1.125=1.78
\n" ); document.write( "area under curve for +-1.78=.9250
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\n" ); document.write( "3)
\n" ); document.write( "1/1.125 = .89
\n" ); document.write( "area under curve for +-0.89=0.6266
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\n" ); document.write( "4)
\n" ); document.write( "For the area under the curve of 0.8 the z=+-1.28
\n" ); document.write( "1.28 * 9/8 = 1.44 ml
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\n" ); document.write( "5)
\n" ); document.write( "For the area under the curve of 0.95 the z=+-1.96
\n" ); document.write( "1.96 * 9/sqrt(x)=1 ml
\n" ); document.write( "sqrt(x)=1.96*9=17.64
\n" ); document.write( "x=311.17=312 always round up.
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\n" ); document.write( "Ed
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