Question 927741: Hello, I am struggling with the following statistical problem:
"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?"
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.
Thank you in advance for your help.
Regards
Nela
Answer by stanbon(75887)   (Show Source):
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?"
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*2=500 attendants + 20% invite 1 guest -> 100*1=100 guests which sums up to 600 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.
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The # that will attend = 600
The # that could attend = 1000
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The proportion that will attend = 0.6
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The standard deviation = sqrt[pq/n] = sqrt[0.6*0.4/1000] = 0.015
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Margin of Error:: z*s = 1.96*0.015 = 0.03
95% CI:: 0.6 - 0.03 < p < 0.6 + 0.03
95%CI:: 0.57 < p < 0.63
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Ans: # of chairs needed with 95% confidence
is between 0.57*1000 and 0.63*1000
or 570 to 630
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Cheers,
Stan H.
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