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 92.2% for the medical students admitted through special programs. Be sure to enter at least 4 digits of accuracy for this problem! If 11 of the students from the special programs are randomly selected, find the probability that at least 10 of them graduated. prob = At least 4 digits! If 11 of the students from the special programs are randomly selected, find the probability that eactly 8 of them graduated. prob = At least 4 digits! Would it be unusual to randomly select 11 students from the special programs and get exactly 8 that graduate? no, it is not unusual yes, it is unusual

Answer 1

Here the graduation rate medical students is 92.2%

i.e, probability of a medical student graduate is 0.922

Let define a random variable X that represent number of medical students graduated.

If we select 11 medical students randomly,

$X\sim Binomial (11,0.922)$

P(X = 1) = (1975) 0.922"(1 – 0.922)11-

We need to find the probability that atleast 10 of them graduated.

$i.e, P(X\geq 10)$

$=P(X=10)+P(X=11)$

(10) 11 ) 0.92210 x 0.0781 + 10 (11 10.022" 0.92211 x 0.078°

$=0.38088731+0.40929848$

$=0.79018579$

$\simeq 0.7902$[ round to four decimal place]

The probability is 0.7902

Here we need to find $P(X=8)$

$P(X=8)$

$=\binom{11}{8}0.922^8\times 0.078^3$

$=0.0408898126$

$\simeq 0.0409$[ round to four decimal place ]

The probability is 0.0409

The mean and standard deviation of X is given by,

$X\sim Binomial (11,0.922)$

$\mu=E(X)=np=11\times 0.922=10.142$

$\sigma=\sqrt{Var(X)}=\sqrt{np(1-p)}=\sqrt{11\times 0.922\times 0.078}=0.889424533$

The usual limit of X is given by,

$\therefore (\mu-2\sigma,\mu+2\sigma)$

$=(10.142-2\times 0.889424533,10.142+2\times 0.889424533)$

$=(8.3631509,11.920849)$

$\simeq (8.36,11.92)$

So, the value of X is 8 lie in this usual interval.

Correct answer::

no, it is not unusual