document.write( "Question 912513: . In a certain Caribbean country, the average monthly income per
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document.write( "family is US$5,600 with a standard deviation of US$3,200. If a
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document.write( "random sample of 100 families is selected, calculate the probability
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document.write( "that the sample mean monthly income is:\r
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document.write( "(a)between US$4,000 and US$6,000\r
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document.write( "(b) less than US$5000
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Algebra.Com's Answer #553837 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! m = 5600, sd = 3200 \n" ); document.write( " \n" ); document.write( "a) P(4000 < x < 6000) = normalcdf(4000,6000,5600,320) Using TI \n" ); document.write( "0r \n" ); document.write( "P(4000 < x < 6000) = P( z < 400/320) - P(z < -1400/320) \n" ); document.write( "b)P(x < 5000) = P(z < 600/320) \n" ); document.write( "compute z, use table etc to find Probability \n" ); document.write( " |