document.write( "Question 1151946: Among Coffee drinkers, men drink a mean of 3.2 cups per day, with a standard deviation of 0.8 cups. Assume the number of coffee dinks per day follows a normal distribution. \r
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document.write( "a. What proportion dink 2 cups per day or more?
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document.write( "B. What proportion drink no more than 4 cups per day?
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document.write( "c. If the top 5% of coffee drinkers are considered heavy coffee drinkers, what is the minimum number of ups per day consumed by a heavy coffee drinker? Hint find the 95th percentile.
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Algebra.Com's Answer #773810 by rothauserc(4718)![]() ![]() You can put this solution on YOUR website! mean is 3.2 cups per day and standard deviation is 0.8 cups \n" ); document.write( ": \n" ); document.write( "a. z-score(2) = (2 - 3.2)/0.8 = -1.5 \n" ); document.write( ": \n" ); document.write( "probability(P) associated with a z-score of -1.5 is 0.0668 \n" ); document.write( ": \n" ); document.write( "P (men who drink more than 2 cups per day) = 1 - 0.0668 = 0.9332 is approximately 0.93 \n" ); document.write( ": \n" ); document.write( "Proportion of men who drink more than 2 cups daily is 93/100 \n" ); document.write( ": \n" ); document.write( "b. z-score(4) = (4 - 3.2)/0.8 = 1.0 \n" ); document.write( ": \n" ); document.write( "P associated with a z-score of 1.0 is 0.8413 is approximately 0.84 \n" ); document.write( ": \n" ); document.write( "Proportion of men who drink less than 4 cups a day = 84/100 = 21/25 \n" ); document.write( ": \n" ); document.write( "c. In the table of z-values, a z-score associated with a P of 0.95 is 1.64 \n" ); document.write( ": \n" ); document.write( "(X - 3.2)/0.8 = 1.64 \n" ); document.write( ": \n" ); document.write( "X - 3.2 = 1.64 * 0.8 = 1.312 \n" ); document.write( ": \n" ); document.write( "X = 4.512 is approximately 5 \n" ); document.write( ": \n" ); document.write( "Minimum number of cups consumed by a heavy coffee drinker is 5 cups \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " |