We will solve both parts of the problem step by step.

Questions 1–3: Binomial Distribution

We have a binomial random variable since:

• Each passenger's response is independent.
• There are only two possible outcomes (satisfied or not satisfied).
• The probability of success (satisfaction) is constant at p = 0.90.
• The number of trials is n = 50.

For solving specific probability questions in a binomial setting, we use the formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

Let me know what specific probability you need for these questions!

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Question 4: Poisson Distribution

The number of defects follows a Poisson distribution with:

$\lambda = 1.2 \text{ (mean number of defects per 100 meters)}$

The probability mass function (PMF) of a Poisson-distributed variable is:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

For zero defects (i.e., $X = 0$):

$P(X = 0) = \frac{e^{-1.2} (1.2)^0}{0!}$

$P(X = 0) = e^{-1.2}$

Now, we compute $e^{-1.2}$ and round to two decimal places. math

Given lambda

lambda_val = 1.2

Compute P(X = 0)

P_X_0 = math.exp(-lambda_val)

Round to two decimal places

P_X_0_rounded = round(P_X_0, 2) P_X_0_roundedThe probability that a randomly selected 100-meter fiber optic cable will have zero defects is 0.30 (rounded to two decimal places).

Do you want more details or have any questions?

Here are five related questions you might find useful:

1. What is the probability that exactly 2 passengers are not satisfied with the airline’s service?
2. What is the probability that at least 45 out of 50 passengers are satisfied?
3. What is the probability that a randomly selected fiber optic cable has exactly 2 defects?
4. How does the Poisson distribution approximate a binomial distribution?
5. If the defect rate increases to 2.0 per 100 meters, how does the probability of zero defects change?

Tip: In Poisson distributions, lower values of $\lambda$ increase the probability of zero occurrences, while higher $\lambda$ shifts probability toward nonzero values.