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 91.3% for the medical students admitted through special programs. If 10 of the students from the special programs are randomly selected, find the probability that at least 9 of them graduated.

Answer 1

Answer:

$P(X\geq 9)=P(X=9)+P(X=10)$

And using the probability mass function we got:

$P(X=9)=(10C9)(0.913)^9 (1-0.913)^{10-9}=0.383$  

$P(X=10)=(10C10)(0.913)^{10} (1-0.913)^{10-10}=0.402$  

And replacing we got:

$P(X \geq 9) = 0.383 +0.402= 0.785$

Step-by-step explanation:

Let X the random variable of interest "number of  students graduated", on this case we now that:  

$X \sim Binom(n=10, p=0.913)$  

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

We want to find the following probability:

$P(X\geq 9)=P(X=9)+P(X=10)$

And using the probability mass function we got:

$P(X=9)=(10C9)(0.913)^9 (1-0.913)^{10-9}=0.383$  

$P(X=10)=(10C10)(0.913)^{10} (1-0.913)^{10-10}=0.402$  

And replacing we got:

$P(X \geq 9) = 0.383 +0.402= 0.785$