Question

using the data from the table, what is p(3)? what is the mean of the probability distribution? table: x values (0, 1, 2, 3, 4) and probability: p(x) values (0.1, 0.2, 0.4, 0.2, 0.1)

Explanation:

Step1: Recall the formula for the mean of a probability distribution

The mean (expected value) \( \mu \) of a discrete probability distribution is given by \( \mu=\sum_{i} x_i \cdot P(x_i) \), where \( x_i \) are the values of the random variable and \( P(x_i) \) are their corresponding probabilities.

Step2: Identify the values of \( x \) and \( P(x) \)

From the table:

• When \( x = 0 \), \( P(0)=0.1 \)
• When \( x = 1 \), \( P(1)=0.2 \)
• When \( x = 2 \), \( P(2)=0.4 \)
• When \( x = 3 \), \( P(3)=0.2 \)
• When \( x = 4 \), \( P(4)=0.1 \)

Step3: Calculate each \( x_i \cdot P(x_i) \)

• For \( x = 0 \): \( 0\times0.1 = 0 \)
• For \( x = 1 \): \( 1\times0.2 = 0.2 \)
• For \( x = 2 \): \( 2\times0.4 = 0.8 \)
• For \( x = 3 \): \( 3\times0.2 = 0.6 \)
• For \( x = 4 \): \( 4\times0.1 = 0.4 \)

Step4: Sum the products

Sum all the \( x_i \cdot P(x_i) \) values: \( 0 + 0.2+0.8 + 0.6+0.4=\sum_{i} x_i \cdot P(x_i) \)
First, \( 0.2 + 0.8=1 \), then \( 0.6+0.4 = 1 \), and finally \( 1 + 1=2.1 \)

Answer:

2.1