Answer:
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
