A study was conducted to determine whether there were significant differences between medical students admitted through special programs (such as retention incentive and guaranteed placement programs) and medical students admitted through the regular admissions criteria. It was found that the graduation rate was 92.4% for the medical students admitted through special programs. Be sure to enter at least 4 digits of accuracy for this problem! If 12 of the students from the special programs are randomly selected, find the probability that at least 11 of them graduated.
prob =
At least 4 digits!
If 12 of the students from the special programs are randomly selected, find the probability that exactly 9 of them graduated.
prob =
At least 4 digits!
Would it be unusual to randomly select 12 students from the special programs and get exactly 9 that graduate?
no, it is not unusual
yes, it is unusual
If 12 of the students from the special programs are randomly selected, find the probability that at most 9 of them graduated.
prob =
At least 4 digits!
Would it be unusual to randomly select 12 students from the special programs and get at most 9 that graduate?
yes, it is unusual
no, it is not unusual
Would it be unusual to randomly select 12 students from the special programs and get only 9 that graduate?
no, it is not unusual
yes, it is unusual

A) 0.7696

B) 0.0474

C) Yes it's unusual

D) 0.05746

E) No, it is not unusual

F) No, it is not unusual

Step-by-step explanation:

This is a binomial probability distribution question.

We are told that 92.4% of those admitted graduated.

Thus; p = 92.4% = 0.924

From binomial probability distribution, q = 1 - p

Thus;

q = 1 - 0.924

q = 0.076

Formula for binomial probability distribution is;

P(x) = nCx × p^(x) × q^(n - x)

A) At least 11 graduated out of 12.

P(x ≥ 11) = P(11) + P(12)

P(11) = 12C11 × 0.924^(11) × 0.076^(12 - 11)

P(11) = 0.3823

P(12) = 12C12 × 0.924^(12) × 0.076^(12 - 12)

P(12) = 0.3873

P(x ≥ 11) = 0.3823 + 0.3873

P(x ≥ 11) = 0.7696

B) that exactly 9 of them graduated out of 12. This is;

P(9) = 12C9 × 0.924^(9) × 0.076^(12 - 9)

P(9) = 0.0474

C) We are not given significance level here but generally when not given we adopt a significance level of α = 0.05.

Now, exactly 9 out of 12 that graduated which is P(9) = 0.0474.

We see that 0.0474 is less than the significance level of 0.05. Thus, we can say that it is unusual to randomly select 12 students from the special programs and get exactly 9 that graduate

D) that at most 9 of them out of 12 graduated.

P(x ≤ 9) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9)

This is going to be very long so I will make use of an online probability calculator to get the values of P(0) to P(8) since I already have P(9) as 0.0474.

Thus, we have;

P(0) = 0

P(1) = 0

P(2) = 0

P(3) = 0.00000001468

P(4) = 0.00000040161

P(5) = 0.00000781232

P(6) = 0.00011081163

P(7) = 0.00115477385

P(8) = 0.00877476184

Thus;

P(x ≤ 9) = 0 + 0 + 0 + 0.00000001468 + 0.00000040161 + 0.00000781232 + 0.00011081163 + 0.00115477385 + 0.00877476184 + 0.04741450256

P(x ≤ 9) = 0.05746

E) P(x ≤ 9) = 0.05746 is more than the significance level of 0.05, thus we will say it is not unusual.

F) from online binomial probability calculator, probability of getting only 9 out of 12 is more than the significance value of 0.05. Thus, we will say it is not unusual

hiking down a winding trail.

step-by-step explanation: