Question

Use the following information to answer questions 1-2.
A random variable is used to identify the number of students who attend live lessons in Statistics class for a semester.

X P(x)
6 0.08
7. 0.11
8 0.17
9 0.26
10 0.21
11 0.11
12 0.06

1. Does this table describe a probability distribution? What are the 2 requirements that must be met if it is a probability distribution?

2. What is the mean of the probability distribution in the table above?

Answer 1

1. To determine if the table describes a probability distribution, we need to check two requirements:

a. **Non-negativity**: The probabilities must be between 0 and 1, inclusive. In the provided table, all probabilities (P(x)) are positive and less than or equal to 1, which satisfies this requirement.

b. **Sum of probabilities**: The sum of all probabilities must equal 1.

Let's check the sum:

\[
0.08 + 0.11 + 0.17 + 0.26 + 0.21 + 0.11 + 0.06 = 1.00
\]

Since the sum equals 1, this requirement is also satisfied.

Therefore, the table does describe a probability distribution.

2. To find the mean (expected value) of the probability distribution, we use the formula:

\[
E(X) = \sum (X \cdot P(X))
\]

Now, we calculate the expected value using the values from the table:

\[
E(X) = (6 \cdot 0.08) + (7 \cdot 0.11) + (8 \cdot 0.17) + (9 \cdot 0.26) + (10 \cdot 0.21) + (11 \cdot 0.11) + (12 \cdot 0.06)
\]

Now calculating each term:

- \(6 \cdot 0.08 = 0.48\)
- \(7 \cdot 0.11 = 0.77\)
- \(8 \cdot 0.17 = 1.36\)
- \(9 \cdot 0.26 = 2.34\)
- \(10 \cdot 0.21 = 2.10\)
- \(11 \cdot 0.11 = 1.21\)
- \(12 \cdot 0.06 = 0.72\)

Now, summing these values:

\[
E(X) = 0.48 + 0.77 + 1.36 + 2.34 + 2.10 + 1.21 + 0.72 = 8.08
\]

Thus, the mean of the probability distribution is **8.08**.