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

Question

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?

Answer 1

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.

Answer 2

a) 25C1 * (.02) * (.98)^24 = p(exactly 1 student)

b) 1 - p(no students received accomodation)

c) 1 - p(0 students) - p(1 student)

d) 25 is close to 30, so you can use standard distribution table to look up that value. 2 * (value at 2 standard dev - .5)

3) [(.98)*3 + (.02)*4]/25

Answer 3

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