document.write( "Question 927741: Hello, I am struggling with the following statistical problem:\r
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document.write( "\"Assume that 500 students at a certain college will graduate on a given day. Because of space limitations the college offers each student 2 tickets for the commencement ceremony. From past experience it is know that 50% of the students will invite two guests to attend the ceremony, 20% students will invite one guest, and the remaining 30% will not attend at all, so they will invite no guests. How many chairs should the college order to have at least 95% chance that all attending guests will have seats?\"\r
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document.write( "I thought that we might be able to compute the exact number of attending guests based on the past experience (50% invite 2 guest -> 250*3=750 attendants + 20% invite 1 guest -> 100*2=200 guests which sums up to 950 people attending), so that the corresponding number of chairs could be ordered yielding probability 1 that everyone will get seated. I have no clue though, how to compute the 95% chance of everyone having seats.\r
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document.write( "Thank you in advance for your help.
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document.write( "Regards
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document.write( "Nela \n" );
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Algebra.Com's Answer #563243 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Assume that 500 students at a certain college will graduate on a given day. Because of space limitations the college offers each student 2 tickets for the commencement ceremony. From past experience it is know that 50% of the students will invite two guests to attend the ceremony, 20% students will invite one guest, and the remaining 30% will not attend at all, so they will invite no guests. How many chairs should the college order to have at least 95% chance that all attending guests will have seats?\" \n" ); document.write( " I thought that we might be able to compute the exact number of attending guests based on the past experience \n" ); document.write( "(50% invite 2 guest -> 250*2=500 attendants + 20% invite 1 guest -> 100*1=100 guests which sums up to 600 people attending), \n" ); document.write( "so that the corresponding number of chairs could be ordered yielding probability 1 that everyone will get seated. I have no clue though, how to compute the 95% chance of everyone having seats. \n" ); document.write( "------- \n" ); document.write( "The # that will attend = 600 \n" ); document.write( "The # that could attend = 1000 \n" ); document.write( "------ \n" ); document.write( "The proportion that will attend = 0.6 \n" ); document.write( "---] \n" ); document.write( "The standard deviation = sqrt[pq/n] = sqrt[0.6*0.4/1000] = 0.015 \n" ); document.write( "----- \n" ); document.write( "Margin of Error:: z*s = 1.96*0.015 = 0.03 \n" ); document.write( "95% CI:: 0.6 - 0.03 < p < 0.6 + 0.03 \n" ); document.write( "95%CI:: 0.57 < p < 0.63 \n" ); document.write( "----- \n" ); document.write( "Ans: # of chairs needed with 95% confidence \n" ); document.write( "is between 0.57*1000 and 0.63*1000 \n" ); document.write( "or 570 to 630 \n" ); document.write( "--------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "-------------- \n" ); document.write( " |