Question

In: Statistics and Probability

A university found that 10% of students withdraw from a math course. Assume 25 students are...

  1. A university found that 10% of students withdraw from a math course. Assume 25 students are enrolled.
    1. Write the prob. distribution function with the specific parameters for this problem

    1. Compute by hand (can use calculator but show some work) the pro that exactly 2 withdraw.
    1. Compute by hand (can use calculator but show some work) the prob. that exactly 5 withdraw.
    1. Construct the probability distribution table of class withdrawals in Excel. This is a table of x and f(x). Generate one column for the number of x successes with numbers 0 to 25 in each row. Generate a second column with the probability f(x) of each success using the Excel function =BINOM.DIST(x, n, p, FALSE) in each row. Attach the table
    1. What is the probability that 5 or less will withdraw?

Solutions

Expert Solution

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


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