Question 841623
A coin is weighted so that heads is twice as likely to appear as tails.
P(heads) = 3/4 ; p(tails) = 1/3
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a) What is the probability that one will get more than five heads if ten such coins are tossed?
Ans: P(6<= x <=10) = 1 - binomcdf(10,3/4,5) = 0.9219
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b)In a school, students are admitted through an interview scores are assumed to follow a normal distribution as X->N (45,100)
I)find the probability that a randomly selected student will score between 40 and 60
z(40) = (40-45)/100 = -5/100 = -1/20
z(60) = (60-45)/100 = 15/100 = 3/20
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P(40<= x <= 60) = P(-1/20<= z <=3/20) = normalcdf(-1/20,3/20) = 0.0796
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ii)If 100 students take the test and the school needs to take 40 students,what should the pass mark be?
z(60) = 3/20
score = (3/20)*100+45 = 60
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iii)explain the limitation of using these statistical limitations for admitting students.
I'll leave that to you.
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Cheers,
Stan H.