The College Board reports that 2% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test. 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 25 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 25 selected students to be?
Answer Step 1 of 6 Given n=25 p=2%=0.02 ,q=1-p=1-0.02=0.98 Let x= no of students receive special accommodations Here X~B(25,0.02) x n-x The pmf of binomial distribution is P(x)=(nc)(p )(q ), x=0,1,2……..,n Step 2 of 6 a) The probability that exactly 1 received a special accommodation =P(x=1) 1 24 =(25c)(0.02) (0.98 )=0.301 Step 3 of 6 b)The probability that at least 1 received a special accommodation =P(x1) =1-P(x=0) =1-0.6035 =0.3965