Question

1. given the probability model in the table below, what is the expected value of the random variable?

x	50	20	5

options: 14, 7.5, 4.67, 5

Explanation:

Step1: Recall the formula for expected value

The expected value \( E(X) \) of a discrete random variable is calculated as \( E(X)=\sum_{i} x_i P(x_i) \), where \( x_i \) are the values of the random variable and \( P(x_i) \) are their corresponding probabilities.

Step2: Calculate each term \( x_i P(x_i) \)

• For \( x = 50 \) and \( P(X)=0.1 \): \( 50\times0.1 = 5 \)
• For \( x = 20 \) and \( P(X)=0.3 \): \( 20\times0.3 = 6 \)
• For \( x = 5 \) and \( P(X)=0.6 \): \( 5\times0.6 = 3 \)

Step3: Sum the terms

Add the results from Step 2: \( 5 + 6 + 3 = 14 \)? Wait, no, wait. Wait, 5 (from 500.1) + 6 (from 200.3) + 3 (from 50.6) = 5 + 6 + 3 = 14? Wait, no, that can't be right. Wait, 500.1 is 5, 200.3 is 6, 50.6 is 3. Then 5 + 6 + 3 = 14? Wait, but let's check again. Wait, 500.1 is 5, 200.3 is 6, 50.6 is 3. So 5 + 6 is 11, plus 3 is 14. Wait, but the options include 14. Wait, maybe I miscalculated earlier. Wait, let's do it again. \( E(X) = 50\times0.1 + 20\times0.3 + 5\times0.6 \). Calculate each product: 500.1 = 5, 200.3 = 6, 50.6 = 3. Then sum: 5 + 6 + 3 = 14.

Answer:

14