To find the mean (expected value) of the given probability distribution, we will use the formula for the mean of a discrete probability distribution:

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

Where:

• $x$ represents the number of defective computers.
• $P(x)$ represents the probability of that number of defective computers occurring.

The given values are:

• $x = 0, P(0) = 0.4096$
• $x = 1, P(1) = 0.4096$
• $x = 2, P(2) = 0.1536$
• $x = 3, P(3) = 0.0256$
• $x = 4, P(4) = 0.0016$

Now, we calculate the mean:

$\mu = (0 \cdot 0.4096) + (1 \cdot 0.4096) + (2 \cdot 0.1536) + (3 \cdot 0.0256) + (4 \cdot 0.0016)$

Breaking this down:

$\mu = 0 + 0.4096 + 0.3072 + 0.0768 + 0.0064$

$\mu = 0.8$

Thus, the mean (expected value) is:

$\mu = 0.80$

The correct answer is μ = 0.80.

Would you like further details or explanations on how to solve this type of problem?

Here are 5 related questions for further exploration:

1. What is the variance of the given probability distribution?
2. How would you calculate the standard deviation of this distribution?
3. What is the interpretation of the mean in a probability distribution?
4. Can you calculate the probability of having at least 2 defective computers?
5. How would the mean change if the probabilities were adjusted?

Tip: The mean of a probability distribution represents the long-run average outcome of a random variable over many trials.