contestada

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 30 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 30 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 30 selected students to be? (Round your answer to two decimal places.)

Respuesta :

Answer:

Step-by-step explanation:

  • Given sample size n = 30 and p = 0.04
  • The application of binomial probability comes in;
  • from binomial probability ; (nCx)px(1−p)(n-x)
  • 1 - p = 0.96

a)  probability that exactly 1 received a special accommodation = P(x=1)

= 30C1 x (0.04) x (0.96) ^29 = 0.3673

b) probability that at least 1 received a special accommodation = P(x >=1) = 1 -P(x = 0)

= 1 - [ 30C0 x (0.04)^0 x (0.96)^30 = 0.7061

c) probability that at least 2 received a special accommodation

= P(x>=2) = 1 - [P(x = 0) + P(x =1)]

= 1 - [ 30C0 x (0.04)^0 x (0.96)^30 + 30C1 x (0.04)^1 x (0.96)^29

= 0.3388

d) probability that the number among the 30 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated ;

= P( x<=1) = P( x= 0) = 0.2939

e) expected average time = 0.04 x 4.5 + 0.96 x 3 = 3.06hours

Otras preguntas