Find the mean of the given probability distribution.

The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4096, 0.4096, 0.1536, 0.0256, and 0.0016, respectively. Round answer to the nearest hundredth.
Group of answer choices
μ = 0.80
μ = 2.00
μ = 0.70
μ = 1.21

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.

Find the mean of the given probability distribution. The probabilities that a batch of 4 computers will contain 0, 1, 2, 3, and 4 defective computers are 0.4096, 0.4096, 0.1536, 0.0256, and 0.0016, respectively. Round answer to the nearest hundredth.

Learn how to calculate the mean (expected value) of a discrete probability distribution, where the probabilities of 0, 1, 2, 3, and 4 defective computers in a batch are given. The solution uses the formula μ = Σ(x * P(x)) and is suitable for students in Grades 10-12.

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