Question 203364: Please help me solve the following question. It is from a former Statistics Question paper of my university and I am unable to find the solution.
Question: PART A; It is observed in a student coffee lounge that only 15% students have to wait for more than five minutes to be served.
Calculate the following probabilities relating to a random sample of 6 students.
1. The probability that none of the 6 students have to wait for more than five minutes.
2. The probability that fewer than 2 students wait for more than five minutes.
3. The probability that either one or two students wait for more than five minutes.
PART B: If we wish to calculate probabilities relating to 100 students a new method is needed. Suggest and explain a suitable methond and explain the term 'Continuity correction.'
USE your method to calculate the following;
1. The probability that exactly 15 out of 100 wait for more than 5 minutes.
2. The probability that the number of students waiting for more than 5 minutes lies between 12 and 16.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Question: PART A; It is observed in a student coffee lounge that only 15% students have to wait for more than five minutes to be served.
Calculate the following probabilities relating to a random sample of 6 students.
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P(wait more than 5 min) = 0.15
P(wait less than 5 min) = 0.85
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1. The probability that none of the 6 students have to wait for more than five minutes.
Ans: (0.85)^6
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2. The probability that fewer than 2 students wait for more than five minutes.
Ans: P(0
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3. The probability that either one or two students wait for more than five minutes.
Ans: P(x = 1 or x = 2)) = 0.5755
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PART B: If we wish to calculate probabilities relating to 100 students a new method is needed. Suggest and explain a suitable method and explain the term 'Continuity correction.'
USE your method to calculate the following;
1. The probability that exactly 15 out of 100 wait for more than 5 minutes.
Using the "normal approximation" you calculate the probability that
x is between 14.5 and 15.5 using a mean of np = 100*0.15 = 15 and a
standard devistion of sqrt(npq)=sqrt(15*0.85) = 3.5707
Ans: P(x=15) is approx normalcdf(14.5,15.5,15,3.5707) = 0.1114
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2. The probability that the number of students waiting for more than 5 minutes lies between 12 and 16.
Ans: normalcdf(11.5,16.5,15,3.5707) = 0.4993
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Cheers,
Stan H.
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