Question

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 89.6% for the medical students admitted through special programs. If 9 of the students from the special programs are randomly selected, find the probability that at least 8 of them graduated. round your answer to 4 decimal places

Answer 1

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;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

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%

LET X = Number of graduated medical students who had admitted through special programs

So, it means X ~ Binom(n = 9, p = 0.896)

Now, probability that at least 8 of them graduated is given by = P(X $\geq$ 8)

    P(X $\geq$ 8) = P(X = 8) + P(X = 9)

                  =  $\binom{9}{8}\times 0.896^{8} \times (1-0.896)^{9-8}+ \binom{9}{9}\times 0.896^{9} \times (1-0.896)^{9-9}$

                  =  $9 \times 0.896^{8} \times 0.104^{1} +1 \times 0.896^{9} \times 1$

                  =  0.7610

Therefore, the probability that at least 8 of them graduated is 0.7610.