Let’s break down each part of this problem.

Problem (a): Exponential Distribution

We have:

• A system with components whose time to failure $F$ follows an exponential distribution with mean $\beta = 5$ years.
• There are 5 independent components installed, and we are asked to find the probability that at least 2 of them are still functioning after 8 years.

Step 1: Probability of a Single Component Functioning for 8 Years

The exponential distribution has a probability density function $f(t) = \frac{1}{\beta} e^{-t/\beta}$ and a cumulative distribution function $F(t) = 1 - e^{-t/\beta}$.

1. The probability that a component is still functioning after 8 years (i.e., $P(T > 8)$) is: $P(T > 8) = 1 - F(8) = e^{-8/\beta} = e^{-8/5}$

2. Calculating $e^{-8/5}$: $P(T > 8) \approx e^{-1.6} \approx 0.2019$

Step 2: Probability That at Least 2 Components Are Functioning

Let $X$ represent the number of components functioning after 8 years. $X$ follows a binomial distribution, $X \sim \text{Binomial}(n = 5, p = 0.2019)$, where $p = 0.2019$ is the probability of a single component still working after 8 years.

We need $P(X \geq 2)$, which is: $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$

Using the binomial probability formula $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$:

1. Calculate $P(X = 0)$: $P(X = 0) = \binom{5}{0} (0.2019)^0 (1 - 0.2019)^5 = (0.7981)^5 \approx 0.3277$

2. Calculate $P(X = 1)$: $P(X = 1) = \binom{5}{1} (0.2019)^1 (0.7981)^4 \approx 5 \times 0.2019 \times 0.4066 \approx 0.4105$

3. Calculate $P(X \geq 2)$: $P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \approx 1 - 0.3277 - 0.4105 \approx 0.2618$

Thus, the probability that at least 2 components are still functioning after 8 years is approximately 0.2618.

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Problem (b): Gamma Distribution for Complaint Time

For this part:

• The time between customer complaints follows a gamma distribution with parameters $\alpha = 2$ and $\beta = 4$.
• Following quality control changes, 20 months passed before a complaint.
• We need to determine if this 20-month interval suggests an effective change.

1. Expected Time Between Complaints (Mean): For a gamma distribution, the mean $\mu = \alpha \cdot \beta = 2 \cdot 4 = 8$ months.

2. Probability of Observing a Gap of 20 Months or More: We calculate $P(X \geq 20)$ using the cumulative distribution function (CDF) for the gamma distribution.

Since $P(X \leq 5) = 0.96$ is given, we can infer that 5 months is a very short interval relative to typical gaps, while 20 months would be unusually long, suggesting that quality control was likely effective.

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Would you like to go deeper into each calculation or have further clarifications?

Further Questions to Explore:

1. How would the result in part (a) change if we increased the mean time to failure?
2. What happens to the probability in part (a) if the components are not independent?
3. How does a gamma distribution differ from an exponential distribution, and when is each used?
4. Can we quantify how "effective" quality control was with more data?
5. How do the parameters $\alpha$ and $\beta$ influence the shape of the gamma distribution?

Tip:

Gamma and exponential distributions are related: the exponential distribution is a special case of the gamma distribution when $\alpha = 1$.