document.write( "Question 624406: Please show me specific detailed answers to these question i got wrong on my test. Thankyou. \r
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document.write( "An internal survey of a company showed that their staff needs to spend about 55 minutes from home to work every morning on average. It was also reported that 3% of the staff needs over two hours for the travel.\r
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document.write( "(i) If 50 staff is surveyed on a particular day, what is the probability that 5 will report the travel of over two hours?\r
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document.write( "(ii) If the company has 3000 staff, what is the expected number of staff who has the travel of over two hours?\r
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document.write( "(iii) If the company has 3000 staff, what are the variance and standard deviation of the number of staff who has the travel of over two hours?
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Algebra.Com's Answer #392770 by Theo(13342)![]() ![]() You can put this solution on YOUR website! 55 minutes is the average. \n" ); document.write( "120 minutes is equal to 2 hours. \n" ); document.write( "3% take over 120 minutes. \n" ); document.write( "this looks like a binomial distribution type of problem. \n" ); document.write( "binomial distribution formula is: \n" ); document.write( "p(x) out of n = C(n,x) * p^x * q^(n-x) \n" ); document.write( "p(x) = probability of number of events that are successful \n" ); document.write( "n = total number \n" ); document.write( "p = probability that commute time will be greater than 120 minutes (success) \n" ); document.write( "q = probability that commute time will be less than or equal to 120 minutes (failure) \n" ); document.write( "c(n,x) is the combination formula for x events out of n = n! / (x! * (n-x)!)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "based on that:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i) If 50 staff are surveyed on a particular day, what is the probability that 5 will report the travel of over two hours?\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "p(5) out of 50 = C(50,5) * (.03)^5 * (.97)^45 = .013074229\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(ii) If the company has 3000 staff, what is the expected number of staff who has the travel of over two hours?\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "expected number would be equal to n*p = 50 * .03 = 90\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "(iii) If the company has 3000 staff, what are the variance and standard deviation of the number of staff who has the travel of over two hours?\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "based on the binomial distribution formula:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "variance is equal to n*p*q = 3000 * .03 * .97 = 87.3 \n" ); document.write( "standard deviation is equal to square root of variance is equal to 9.343446901\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "a reference is here: \n" ); document.write( "http://www.stat.yale.edu/Courses/1997-98/101/binom.htm\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i think these are the answers you are looking for but i'm not totally sure. \n" ); document.write( "check them out and see if they agree with the answers you were supposed to have.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |