Let's start by calculating the expected value (mean) for Distribution A. The formula for the expected value is:

$\mu = \sum (x_i \cdot P(X = x_i))$

For Distribution A, the values for $x_i$ and $P(X = x_i)$ are as follows:

$x_i$	$P(X = x_i)$
0	0.06
1	0.12
2	0.17
3	0.25
4	0.40

Now we calculate the expected value for Distribution A:

$\mu_A = (0 \cdot 0.06) + (1 \cdot 0.12) + (2 \cdot 0.17) + (3 \cdot 0.25) + (4 \cdot 0.40)$

Let's compute this:

$\mu_A = (0) + (0.12) + (0.34) + (0.75) + (1.60) = 2.81$

So, the expected value for Distribution A is:

$\mu_A = 2.81$

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Now, let's calculate the expected value for Distribution B using the same formula:

$x_i$	$P(X = x_i)$
0	0.40
1	0.25
2	0.17
3	0.12
4	0.06

$\mu_B = (0 \cdot 0.40) + (1 \cdot 0.25) + (2 \cdot 0.17) + (3 \cdot 0.12) + (4 \cdot 0.06)$

Let's compute this:

$\mu_B = (0) + (0.25) + (0.34) + (0.36) + (0.24) = 1.19$

So, the expected value for Distribution B is:

$\mu_B = 1.19$

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Answers:

• a. Expected value for Distribution A: $\mu_A = 2.81$
• b. Expected value for Distribution B: $\mu_B = 1.19$

Do you want me to proceed with the standard deviation, or do you have any questions so far?

Here are five related questions that expand on this information:

1. How is the variance related to the standard deviation in a probability distribution?
2. How can expected values be interpreted in real-life contexts?
3. What happens to the expected value if all probabilities are equal?
4. What is the significance of a higher expected value in a probability distribution?
5. How do you compute the expected value for a continuous distribution?

Tip: The expected value is often referred to as the long-run average or mean when considering repeated trials of a probability experiment.