The problem involves a robot that paints cars, with a known error rate. Specifically:

• The robot completes 81% of its work without errors, meaning 19% are defective.
• The random variable $X$ represents the number of error-free jobs out of a random sample of $n = 8$ cars.

Part a)

It asked for the probability that the number of error-free jobs is 6 or greater. This seems to have already been solved, yielding a correct result of: $P(X \geq 6) = 0.285 + 0.348 + 0.185 = 0.818.$

Part b)

Here, you're asked to compute the expected number of error-free jobs, $E[X]$.

Since $X$ follows a binomial distribution with parameters $n = 8$ and the probability of success $p = 0.81$, the formula for the expectation is: $E[X] = n \cdot p.$

Substituting the values: $E[X] = 8 \cdot 0.81 = 6.48.$

So, the expected number of error-free jobs is 6.480.

Part c)

Next, you are asked to compute the variance of the number of error-free jobs, $\text{Var}(X)$.

For a binomial distribution, the variance is given by the formula: $\text{Var}(X) = n \cdot p \cdot (1 - p).$

Substituting the values: $\text{Var}(X) = 8 \cdot 0.81 \cdot (1 - 0.81) = 8 \cdot 0.81 \cdot 0.19 = 1.2288.$

So, the variance is approximately 1.229.

Summary:

• b) $E[X] = 6.480$
• c) $\text{Var}(X) = 1.229$

Would you like more details on any of these steps?

Here are 5 related questions to explore further:

1. How would the probability change if the robot’s error rate decreased?
2. How would the variance change for different sample sizes?
3. What if the error probability was unknown? How would you estimate it?
4. Can we calculate the probability for exactly 6 error-free jobs instead of 6 or greater?
5. How can you interpret the variance in terms of the robot's performance?

Tip: The binomial distribution is used when there are exactly two possible outcomes (like success/failure), and you can model many real-world processes with it, such as manufacturing or testing.