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

Question

Suppose that 2% 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 20 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 20 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 20selected students to be? (Round your answer to two decimal places.)

Explanation / Answer

Ans:

Binomial distribution:

n=20,p=0.02

a)P(k=1)=20C1*0.021*0.9819

=0.2725

using Eexcel function:

P(k=1)=BINOMDIST(1,20,0.02,FALSE)=0.2725

b)P(k>=1)=1-P(k=0)=1-20C0*0.020*0.9820

=1-0.9820

=1-0.6676

=0.3324

c)P(k>=2)=1-P(k=0)-P(k=1)=1-0.6676-0.2725=0.0599

or use excel function

P(k>=2)1-P(k<=1)=1-BINOMDIST(1,20,0.02,TRUE)=0.0599

d)mean=np=20*0.02=0.4

standard deviation=sqrt(0.4*0.98)=0.63

P(|X µ| 2) = P(X µ + 2) = P(X 0.4 + (2)(0.63)) = P(X 1.66) = P(X = 0) + P(X = 1) 0.6676+ 0.2725 = 0.9401.

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