Question

Let \( X \) be a random variable with the probability mass function 

\[P(X = n) = \begin{cases} \dfrac{1}{10}, & n = 1,2,\dots,10, \\ 0, & \text{otherwise}.\end{cases}\] 

Then \( E(\max\{X, 5\}) \) equals ............ 
 

Hint:

When dealing with functions of random variables, break them down into cases and use the law of total expectation.

Answer 1

Correct Answer: 6.25 - 6.75

Step 1: Define the random variable of interest. 
We are asked to find \( E(\max\{X, 5\}) \). The function \( \max\{X, 5\} \) means that if \( X \) is greater than or equal to 5, then \( \max\{X, 5\} = X \), and if \( X \) is less than 5, then \( \max\{X, 5\} = 5 \).

Step 2: Compute the expected value. 
We compute the expected value of \( \max\{X, 5\} \) by considering the probabilities: \[ E(\max\{X, 5\}) = \sum_{n=1}^{10} P(X = n) \cdot \max\{n, 5\}. \] Since \( P(X = n) = \frac{1}{10} \) for \( n = 1, 2, \dots, 10 \), we calculate: \[ E(\max\{X, 5\}) = \frac{1}{10} \left( \sum_{n=1}^{4} 5 + \sum_{n=5}^{10} n \right). \] This simplifies to: \[ E(\max\{X, 5\}) = \frac{1}{10} \left( 4 \times 5 + 5 + 6 + 7 + 8 + 9 + 10 \right) = \frac{1}{10} \times 65 = 6.5. \]

Step 3: Conclusion. 
Thus, \( E(\max\{X, 5\}) = 6.5 \).