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 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? 352 (b) What is the probability that at least 1 received a special accommodation? 845 (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.) hr You may need to use the appropriate table in the Appendix of Tables to answer this question

Explanation / Answer

Solution:-

p = 0.02

n = 35

a) The probability that excatly one recieved a special recommendation is 0.352.

x = 1

By applying binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x = 1) = 0.352

b) The probability that atleast one recieved a special recommendation is 0.5069.

x = 1

By applying binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x > 1) = 0.5069

c) The probability that atleast 2 recieved a special recommendation is 0.1547

x = 2

By applying binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x > 2) = 0.1547

d) The probability that number among the 35 who recieved a special accomodation is within 2 standard deviations is 0.9544.

Within two standard deviation means that z score value = + 2

P(- 2 < z < 2) = P(z > - 2) - P(z > 2)

P(- 2 < z < 2) = 0.9772 - 0.0228

P(- 2 < z < 2) = 0.9544

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