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 |