To find the mean (μ) of a probability distribution, we use the formula:

$\mu = \sum (x \times P(x))$

Where:

• $x$ represents the possible outcomes (in this case, the number of bases).
• $P(x)$ represents the probability of each outcome.

Given the dataTo find the mean (μ) of a probability distribution, we use the formula:

$\mu = \sum (x \times P(x))$

Where:

• $x$ represents the possible outcomes (in this case, the number of bases).
• $P(x)$ represents the probability of each outcome.

Given the

\text{Bases} (x) & \text{Probability} (P(x)) \\ \hline 0 & 0.7442 \\ 1 & 0.1739 \\ 2 & 0.0528 \\ 3 & 0.0002 \\ 4 & 0.0289 \\ \end{array}$$ Let's calculate the mean (μ): $$\mu = (0 \times 0.7442) + (1 \times 0.1739) + (2 \times 0.0528) + (3 \times 0.0002) + (4 \times 0.0289)$$ $$\mu = (0) + (0.1739) + (0.1056) + (0.0006) + (0.1156)$$ $$\mu = 0.3957$$ So, the mean of this probability distribution is **0.3957**. Would you like more details or have any questions? --- **Related Questions:** 1. What is the variance of this probability distribution? 2. How do you calculate the standard deviation of this distribution? 3. How would the mean change if the probabilities were slightly different? 4. What is the interpretation of the mean in the context of this problem? 5. Can the mean be a non-integer value even if the outcomes are integers? 6. How does the mean relate to the median in a probability distribution? 7. What is the probability that the outcome will be greater than the mean? 8. How can the mean be used to predict future outcomes in similar distributions? **Tip:** The mean of a probability distribution is sometimes called the expected value, as it represents the average outcome if the experiment were repeated many times.