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 |