Answer:
Probability that at least 8 of them graduated is 0.7610.
Step-by-step explanation:
We are given that it was found that the graduation rate was 89.6% for the medical students admitted through special programs.
Also, 9 of the students from the special programs are randomly selected.
The above situation can be represented through Binomial distribution;
![P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....](https://tex.z-dn.net/?f=P%28X%3Dr%29%20%3D%20%5Cbinom%7Bn%7D%7Br%7Dp%5E%7Br%7D%20%281-p%29%5E%7Bn-r%7D%20%3B%20x%20%3D%200%2C1%2C2%2C3%2C.....)
where, n = number of trials (samples) taken = 9 students
r = number of success = at least 8
p = probability of success which in our question is % of graduation
rate for the medical students admitted through special programs,
i.e; 89.6%
<em>LET X = Number of graduated medical students who had admitted through special programs</em>
<u>So, it means X ~ Binom(n = 9, p = 0.896)</u>
Now, probability that at least 8 of them graduated is given by = P(X
8)
P(X
8) = P(X = 8) + P(X = 9)
=
=
= <u>0.7610</u>
<em>Therefore, the probability that at least 8 of them graduated is 0.7610.</em>