Let's break down the questions from the image and solve them step by step:

1. Fill in the missing cell in the probability distribution.

To find the missing probability, we know that the sum of all probabilities in a probability distribution must equal 1.

Let's list out the given probabilities:

$P(0) = 0.22,\ P(1) = 0.15,\ P(2) = 0.12,\ P(3) = 0.09,\ P(4) = 0.05,\ P(5) = 0.04,\ P(6) = 0.03,\ P(8) = 0.14$

Let $P(7)$ be the missing probability.

We set up the equation: $0.22 + 0.15 + 0.12 + 0.09 + 0.05 + 0.04 + 0.03 + P(7) + 0.14 = 1$

Simplifying: $0.84 + P(7) = 1$ $P(7) = 1 - 0.84 = 0.16$

Thus, the missing probability $P(7) = 0.16$.

2. What is the probability that a randomly selected container will contain at most 2 broken components?

"At most 2 broken components" means $0, 1,$ or $2$ broken components. We sum the probabilities for these cases:

$P(\text{at most 2}) = P(0) + P(1) + P(2)$ $P(\text{at most 2}) = 0.22 + 0.15 + 0.12 = 0.49$

So, the probability is 0.49.

3. What is the probability that a randomly selected container will contain more than 4 broken components?

"More than 4 broken components" refers to the cases where there are 5, 6, 7, or 8 broken components. We sum the probabilities for these cases:

$P(\text{more than 4}) = P(5) + P(6) + P(7) + P(8)$ $P(\text{more than 4}) = 0.04 + 0.03 + 0.16 + 0.14 = 0.37$

Thus, the probability is 0.37.

4. What is the probability that a randomly selected container will contain at least 3 functioning components?

If a container has at least 3 functioning components, that means there are at most 5 broken components (since there are 8 components in total, $8 - 3 = 5$).

So, the probability of at least 3 functioning components is the probability of having 0 to 5 broken components:

$P(\text{at least 3 functioning}) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5)$ $P(\text{at least 3 functioning}) = 0.22 + 0.15 + 0.12 + 0.09 + 0.05 + 0.04 = 0.67$

Thus, the probability is 0.67.

5. What is the mean of the probability distribution?

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

$\mu = \sum x_i P(x_i)$ where $x_i$ is the number of broken components and $P(x_i)$ is the probability for each number.

Using the given values:

$\mu = 0(0.22) + 1(0.15) + 2(0.12) + 3(0.09) + 4(0.05) + 5(0.04) + 6(0.03) + 7(0.16) + 8(0.14)$

Calculating:

$\mu = 0 + 0.15 + 0.24 + 0.27 + 0.20 + 0.20 + 0.18 + 1.12 + 1.12$ $\mu = 3.48$

Thus, the mean is 3.48 broken components.

Let me know if you want further details or clarifications on any step!

Related Questions:

1. How would you calculate the variance of this probability distribution?
2. What is the probability of having exactly 3 broken components?
3. How does the cumulative probability function look for this distribution?
4. Can you compute the probability of having at least 4 broken components?
5. What is the mode of the probability distribution?

Tip:

When dealing with probability distributions, always verify that the sum of the probabilities equals 1 to ensure the distribution is valid.