A selective college would like to have an entering class of 1000 students. Becau

Question

A selective college would like to have an entering class of 1000 students. Because not all students who are offered admission accept, the college admits more than 1000 students. Past experience shows that about 83% of the students admitted will accept. The college decides to admit 1200 students. Assuming that students make their decisions independently, the number who accept has the B(1200, 0.83) distribution. If this number is less than 1000, the collegel admit students from its waiting list. (a) what are the mean and the standard deviation of the number x of students who accept? (Round your standard deviation to four decimal places.) (b) Use the normal approximation to find the probability that at least 790 students accept. (Round your answer to four decimal places.) (c) The college does not want more than 1000 students. What is the probability that more than 1000 will accept? (Round your answer to four decimal places.) (d) If the college decides to decrease the number of admission offers to 1,150, what is the probability that more than 1000 will accept? (Round your answer to four decimal places.) eBook Submit AnswerSaveProgresel

Explanation / Answer

Ans:

Binomial distribution with n=1200,p=0.83

a)mean=np=1200*0.83=996

standard deviation=sqrt(np(1-p)=sqrt(1200*0.83*(1-0.83)=sqrt(169.32)=13.0123

b)it is normal distribution with mean=996 and standard deviation=13.0123

P(X>=790)

z=(790-996)/13.0123=-15.831

P(z>=-15.831)=1-P(z<=-15.831)=1-0=1

c)

P(X>1000)

z=(1000-996)/13.0123=0.3074

P(z>0.3074)=1-P(z<=0.3074)=1-0.6207=0.3793

d)mean=1150*0.83=954.5

standard deviation=sqrt(1150*0.83*(1-0.83))=12.7383

z=(1000-954.5)/12.7383=3.572

P(z>3.572)=1-P(z<=3.572)=1-0.9998=0.0002

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