Suppose that 4% of the 2 million high school students who take the SAT each year

Question

Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities. Consider a random sample of 35 students who have recently taken the test. (Round your probabilities to three decimal places.)

(a) What is the probability that exactly 1 received a special accommodation?

(b) What is the probability that at least 1 received a special accommodation?

(c) What is the probability that at least 2 received a special accommodation?

(d) What is the probability that the number among the 35 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

(e) Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed 4.5 hours. What would you expect the average time allowed the 35 selected students to be? (Round your answer to two decimal places.)

Please show work

Explanation / Answer

There are two outcomes-success (receive special accomodation) and failure (donot receive special accomodations), and the probability of success is 0.04. There are n=35 random, independent trials. This accounts for binomial distribution. Use P(X,r)=nCr(p)^r(1-p)^n-r, where, r denotes specific number of success in n trials.

a. P(X=1)=35C1(0.04)^1(1-0.04)^34=0.3494

b. P(X>=1)=1-P(X<1)=1-P(X=0)=1-35C0(0.04)^0(1-0.04)^35=0.7604

c. P(X>=2)=1-P(X<2)=1-[P(X=0)+P(X=1)]=1-[0.2396+0.3494]=0.4110

d. The expected number of special accomodation received, mu=np=35*0.04=1.4

Therefore, the number within 2 standard deviations of expected value is 2*1.4=2.8~3

Therefore, required probability is 3/35=0.0857

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.