In: Statistics and Probability
Note:
Hey there! Thank you for the question. As you have posted several subparts, as per our policy, we have solved the first 4 subparts for you.
a.
There are n = 25 students enrolled. Out of these, 10% withdraw from a math course. Thus, the probability of withdrawal is, p = 0.10. It can be assumed that the students withdraw independently.
Hence, the random variable (say, X) representing the number of students who withdraw among the 25, is a binomially distributed random variable.
If X ~ Binomial (n, p), then the probability mass function of X is:

In this case, X has a binomial distribution with parameters n = 25, p = 0.10. Note that, q = 1 – p = 0.90. Thus, X ~ Binomial (25, 0.10). The probability distribution function is:

b.
The probability that exactly 2 students withdraw is, P (X = 2) = p (2), calculated below:

Thus, the probability that exactly 2 students withdraw is 0.2659.
c.
The probability that exactly 5 students withdraw is, P (X = 5) = p (5), calculated below:

Thus, the probability that exactly 2 students withdraw is 0.0646.
d.
By following the instructions given in Part d, the following table is obtained. We have taken the probabilities to 4 decimal places.
| 
 x  | 
 f (x)  | 
| 
 0  | 
 0.0718  | 
| 
 1  | 
 0.1994  | 
| 
 2  | 
 0.2659  | 
| 
 3  | 
 0.2265  | 
| 
 4  | 
 0.1384  | 
| 
 5  | 
 0.0646  | 
| 
 6  | 
 0.0239  | 
| 
 7  | 
 0.0072  | 
| 
 8  | 
 0.0018  | 
| 
 9  | 
 0.0004  | 
| 
 10  | 
 0.0001  | 
| 
 11  | 
 0.0000  | 
| 
 12  | 
 0.0000  | 
| 
 13  | 
 0.0000  | 
| 
 14  | 
 0.0000  | 
| 
 15  | 
 0.0000  | 
| 
 16  | 
 0.0000  | 
| 
 17  | 
 0.0000  | 
| 
 18  | 
 0.0000  | 
| 
 19  | 
 0.0000  | 
| 
 20  | 
 0.0000  | 
| 
 21  | 
 0.0000  | 
| 
 22  | 
 0.0000  | 
| 
 23  | 
 0.0000  | 
| 
 24  | 
 0.0000  | 
| 
 25  | 
 0.0000  | 
| 
 Total  | 
 1  |