document.write( "Question 1193773: 8 Last year, at Northern Manufacturing Company, 200
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document.write( "people had colds during the year. One hundred fiftyfive people who did no exercising had colds, and the
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document.write( "remainder of the people with colds were involved in
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document.write( "a weekly exercise program. Half of the 1,000 employees were involved in some type of exercise.
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document.write( "(a) What is the probability that an employee will
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document.write( "have a cold next year?
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document.write( "(b) Given that an employee is involved in an exercise program, what is the probability that he or
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document.write( "she will get a cold next year?
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document.write( "(c) What is the probability that an employee who is
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document.write( "not involved in an exercise program will get a
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document.write( "cold next year?
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document.write( "(d) Are exercising and getting a cold independent
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document.write( "events? Explain your answer \n" );
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Algebra.Com's Answer #825841 by Boreal(15235) You can put this solution on YOUR website! 500 exercised and 500 didn't, and the total is 1000 \n" ); document.write( "this is what you are given \n" ); document.write( "----------C--------NC------Total \n" ); document.write( "E------- 45-----------------500 \n" ); document.write( "NE------155---------------- 500 \n" ); document.write( "Total---200----------------1000\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "From that, you can fill out the rest of the table\r \n" ); document.write( "\n" ); document.write( "----------C--------NC------Total \n" ); document.write( "E------- 45-------455-------500 \n" ); document.write( "NE------155------345-------500 \n" ); document.write( "Total---200-------800------1000\r \n" ); document.write( "\n" ); document.write( "a. 20% \n" ); document.write( "b. 45/500 or 9% \n" ); document.write( "c. 31%, assuming you can generalize the results to next year \n" ); document.write( "d.If exercise and getting a cold are independent then the probability(E), 0.5, and probability(C), 0.2 equals the probability of P(E and C), which is 0.09. \n" ); document.write( "0.10 NE 0.09, so not independent. They are dependent to some extent. \n" ); document.write( " |