Probabality questions below need help fast is about discrete random varibles and probablity distributions?

the college board reports 2% of the 2 million high school students who take the SAT each yr get special accomadations because of documented disablities. consider a random variable of 25 students who have tkaen the test


a) what is the probablity that exactly 1 recieved special accomadation?


b) what is the probablity that least1 recieved special accomadation?


c) what is the probablity that at least 2 recieved special accomadation?


d)what is the probablity that the number among the 25 who recieve a special accomodation is within 2 deviations of the number you would expect to be accomadated?


e)suppose that a student who does ot recieve a special accodadtion is allowed 3 hrs for the exam, where as an accomodated student is allowed 4.5 hrs. what would you expect the average time allowed the 25 students to be?

Probabality questions below need help fast is about discrete random varibles and probablity distributions?
Let X be the number of student who receive special accommodations. X has the binomial distribution with n = 25 trails and success probability p = 0.02





In general, if X has the binomial distribution with n trials and a success probability of p then





X = x] = 0 for any other value of x.





this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.





a) P(X = 1) = 0.3078902





b) P(X ≥ 1) = 1 - P(X = 0) = 1 - 0.6034647


= 0.3965353





c) P(X ≥ 2) = 1 - (P(X = 0) + P(X = 1))


= 1 - (0.60346473 + 0.30789017)


= 0.0886451





d) the mean of the binomial is n * p = 25 * 0.02 = 0.5


the variance is n * p * (1-p) = 0.49


the standard deviation is the square root of the variance = 0.7





P(0.5 - 2 * 0.7 ≤ X ≤ 0.5 + 0.2 * 0.7)


= P( -0.9 ≤ X ≤ 1.9)


since you have a discrete random variable the fractions are not possible


= P( 0 ≤ X ≤ 1) = P(X = 0) + P(X = 1) = 0.9113549





e) 24.5 students will take the exam in 3 hours, 0.5 students will take the exam in 4.5 hours, the average time is:





(24.5 * 3 + 0.5 * 4.5) / 25 = 3.03 hours





this is a good example of a weighted mean.