In the table provided, $f(x)$ represents the probability mass function (pmf) $p(x)$ of a random variable $X$, which denotes the number of computer malfunctions on a given day. So, you can interpret $f(x) = p(x)$ as follows:

• $x$ is the number of malfunctions (ranging from 0 to 6).
• $f(x)$ gives the probability of having exactly $x$ malfunctions.

For example:

• $f(0) = 0.17$ means there is a 17% chance that there will be 0 malfunctions.
• $f(1) = 0.29$ means there is a 29% chance that there will be 1 malfunction, and so on.

Now, let’s address the individual parts:

(a) Mean ($\mu$): The mean of the probability distribution can be calculated using the formula:

$\mu = \sum_x x \cdot f(x)$

(b) Standard deviation ($\sigma$): The standard deviation is calculated as:

$\sigma = \sqrt{\sum_x (x - \mu)^2 \cdot f(x)}$

(c) At most 2 malfunctions: Find $P(X \leq 2)$:

$P(X \leq 2) = f(0) + f(1) + f(2)$

(d) More than 3 malfunctions: Find $P(X > 3)$:

$P(X > 3) = f(4) + f(5) + f(6)$

(e) Exactly 3 malfunctions: This is simply $f(3)$.

(f) Between 2 and 5 malfunctions (inclusive): Find $P(2 \leq X \leq 5)$:

$P(2 \leq X \leq 5) = f(2) + f(3) + f(4) + f(5)$

Would you like me to calculate any of these values for you?

Here are 5 follow-up questions you might want to consider:

1. What is the process for calculating the mean for a probability distribution?
2. How can we interpret the standard deviation in the context of this malfunction data?
3. How is the cumulative probability calculated for part (c)?
4. Why do we sum probabilities when calculating $P(X \leq 2)$ or $P(2 \leq X \leq 5)$?
5. How could this malfunction probability distribution help in computer maintenance planning?

Tip: The sum of all probabilities in a probability mass function must always equal 1.