Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum

Question

3 months ago

8

Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum

ented 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 15 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.)

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-02-23T12:00:53+03:00

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Let X denote the number of students who get special accommodation.

P(X) = 4%

P(X) = 0.04

Let S represent the sample size, which equals 30.

Let Y indicate the chosen numbers within the sample size.

Y follows a Binomial distribution, Bin(30,0.04).

a. The chance that exactly 1 student received special accommodation:

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 — rounded

b. The probability that at least 1 received special accommodation is calculated as:

This implies P(Y≥1).

Since P(Y=0) + P(Y≥1) = 1, we find that P(Y≥1) = 1 - P(Y=0).

Now calculating P(Y=0):

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^30

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 — rounded

c.

Determining the probability that at least 2 students received special accommodation:

P(Y≥2) = 1 - P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. To find the probability that among the 15 students, the received special accommodation” falls within 2 standard deviations from what we anticipate,

we first compute the standard deviation:

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Average = np = 15 * 0.04 = 0.6.

The range that embodies two standard deviations away from 0.6 is [0, 2.55], meaning we are interested in the probability that 0, 1, or 2 students out of 20 received special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294.

To find P(2):

P(2) = (0.04)² * (1 - 0.04)^(30 - 2)

P(2) = 0.00051.

Thus,

P(0) + P(1) + P(2) = 0.0122 + 0.294 + 0.00051

= 0.30671.

Therefore, it’s 30.671% likely that 0, 1, or 2 students received accommodations.

e.

The anticipated value from part d) is 0.6.

The expected time is calculated as [.6(4.5) + 19.2(3)]/30 = 2.01 hours.