To determine whether the given table represents a probability distribution, we need to verify two key requirements:

1. The sum of all probabilities must equal 1: For a set of probabilities to define a distribution, the total probability of all possible outcomes must add up to 1.
2. Each individual probability must be between 0 and 1 inclusive: Probabilities represent chances of events, so each probability must fall within this range.

Let's evaluate the given table with these criteria in mind:

$x$	$P(x)$
0	0.034
1	0.156
2	0.310
3	0.310
4	0.156
5	0.034

Step 1: Check if the sum of probabilities equals 1

We sum the probabilities: $0.034 + 0.156 + 0.310 + 0.310 + 0.156 + 0.034 = 1.000$ Since the sum is 1, the first condition is satisfied.

Step 2: Check if all probabilities are between 0 and 1 inclusive

We observe that all probabilities ($P(x)$) are within the range $0 \leq P(x) \leq 1$. This condition is also satisfied.

Conclusion:

Since both conditions are satisfied, the table does represent a valid probability distribution.

Answer:

• The correct choice is A: "Yes, the table shows a probability distribution."

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Part 2: Finding the Mean and Standard Deviation

The formulas for the mean $\mu$ and standard deviation $\sigma$ of a probability distribution are as follows:

1. Mean (Expected Value) $\mu$: $\mu = \sum [x \cdot P(x)]$
2. Variance $\sigma^2$: $\sigma^2 = \sum [(x - \mu)^2 \cdot P(x)]$
3. Standard Deviation $\sigma$: $\sigma = \sqrt{\sigma^2}$

Let's calculate the mean $\mu$:

$\mu = (0 \cdot 0.034) + (1 \cdot 0.156) + (2 \cdot 0.310) + (3 \cdot 0.310) + (4 \cdot 0.156) + (5 \cdot 0.034)$ $\mu = 0 + 0.156 + 0.620 + 0.930 + 0.624 + 0.170 = 2.5$

So, the mean is $\mu = 2.5$.

Next, we'll calculate the variance $\sigma^2$: $\sigma^2 = (0 - 2.5)^2 \cdot 0.034 + (1 - 2.5)^2 \cdot 0.156 + (2 - 2.5)^2 \cdot 0.310 + (3 - 2.5)^2 \cdot 0.310 + (4 - 2.5)^2 \cdot 0.156 + (5 - 2.5)^2 \cdot 0.034$ $\sigma^2 = (6.25 \cdot 0.034) + (2.25 \cdot 0.156) + (0.25 \cdot 0.310) + (0.25 \cdot 0.310) + (2.25 \cdot 0.156) + (6.25 \cdot 0.034)$ $\sigma^2 = 0.2125 + 0.351 + 0.0775 + 0.0775 + 0.351 + 0.2125 = 1.282$

Thus, the variance is $\sigma^2 = 1.282$, and the standard deviation is: $\sigma = \sqrt{1.282} \approx 1.132$

Final Answer:

• The mean is $\mu = 2.5$.
• The standard deviation is $\sigma \approx 1.132$.

Would you like further details or have any questions?

Here are 5 related questions you might explore:

1. How does the mean change if the probability distribution shifts?
2. Can we calculate the mode for this distribution?
3. What is the interpretation of standard deviation in a probability distribution?
4. How would a change in one probability affect the overall distribution?
5. How does variance help in understanding the spread of a distribution?

Tip: The mean of a probability distribution is often called the "expected value," representing the average outcome over many trials.