Question

An investor is interested in purchasing an apartment building containing six apartments. The current owner provides the following probability distribution indicating the probability that the given number of apartments will be rented during a given month.


Number of Rented Apartments 0 1 2 3 4 5 6
Probability 5/38 9/38 7/38 2/19 3/38 1/19 4/19

Find the number of apartments the investor could expect to be rented during a given month. (Hint: Find the expected value.)

a) 2.7232 b) 2.7632 c) 2.7432 d) 2.7932 e) 2.7732

Answer 1

Answer:

The correct answer is B. 2.7632

Step-by-step explanation:

Let's find the number of apartments the investor could expect to be rented during a given month, this way:

0 * 5/38 = 0

1 * 9/38 = 9/38

2 * 7/38 = 14/38

3 * 2/19 = 6/19

4 * 3/38 = 12/38

5 * 1/19 = 5/19

6 * 4/19 = 24/19

In consequence,

9/38 + 14/38 + 6/19 + 12/38 + 5/19 + 24/19 =

38 is the Lowest Common Denominator

(9 + 14 + 12 + 12 + 10 + 48)/38 =

105/38 = 2.7632

The correct answer is B. 2.7632

Answer 2

Using the formula for the mean, the number of apartments the investor could expect to be rented during a given month is:

b) 2.7632

The mean of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.

In this problem, the distribution is:

$P(X = 0) = \frac{5}{38}$

$P(X = 1) = \frac{9}{38}$

$P(X = 2) = \frac{7}{38}$

$P(X = 3) = \frac{2}{19}$

$P(X = 4) = \frac{3}{38}$

$P(X = 5) = \frac{1}{19}$

$P(X = 6) = \frac{4}{19}$

Hence, the expected value is:

$E(X) = 0\frac{5}{38} + 1\frac{9}{38} + 2\frac{7}{38} + 3\frac{2}{19} + 4\frac{3}{38} + 5\frac{1}{19} + 6\frac{4}{19} = \frac{9 + 14 + 12 + 12 + 10 + 48}{38} = 2.7632$

Hence, option b is correct.

A similar problem is given at https://brainly.com/question/24855677