Question

Suppose that 3% 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?

Answer 1

Answer: 0.336

Explanation:

Let x be the number of students who take the SAT each year receive special accommodations.

Probability that a student who take the SAT each year receive special accommodations =0.03

Here ,
`X\sim\text{ Bin(n=20, p=0.03)}`

Binomial probability formula :
`P(X=x)^nC_xp^x(1-p)^(n-x)` , where p =probability of each successes , n is total trials.

Then ,

`P(X=1)=^(20)C_1(0.03)^1(1-0.03)^(20-1)\\\\=(20)(0.03)(0.97)^(19)\approx0.336`

Hence, the probability that exactly 1 received a special accommodation is 0.336 .