Problem 2 - Arbeitsbuch Mathematik für Ingenieure

Question

In einem Kronleuchter werden gleichzeitig 10 Glühbirnen eines bestimmten Typs eingeschraubt. Die Lebensdauer einer Glühbirne dieses Typs (in Stunden) lasse sich durch die exponentialverteilte Zufallsvariable mit \(\lambda=5 \cdot 10^{-4}\) angemessen beschreiben. Für die Lebensdauer der einzelnen. Glühbirnen wird eine Unabhängigkeitsannahme getroffen. a) Bestimmen Sie die Wahrscheinlichkeit, dass eine Glühbirne dieses Typs eine Lebensdauer von über 500 Stunden hat. b) Bestimmen Sie die Wahrscheinlichkeit, dass mindestens 8 der 10 Glühbirnen eine Lebensdauer von über 500 Stunden haben.

Step-by-Step Solution

Answer

a) \( P(T > 500) \approx 0.7788 \) b) \( P(X \ge 8) \approx 0.853 \)

Step 1: Define the exponential distribution

The lifetime of a light bulb follows an exponential distribution with the rate parameter \(\lambda = 5 \times 10^{-4}\). The probability density function (PDF) of an exponential distribution is given by: \[ f(t) = \lambda e^{-\lambda t} \]

Step 2: Calculate the probability of one light bulb lasting over 500 hours

We need to find the probability that a single bulb lasts more than 500 hours. This is given by the survival function of the exponential distribution: \[ P(T > 500) = 1 - F(500) = e^{-\lambda \cdot 500} \] Substitute \( \lambda = 5 \times 10^{-4} \): \[ P(T > 500) = e^{-5 \times 10^{-4} \cdot 500} = e^{-0.25} \]

Step 3: Compute the value of the exponential expression

Calculate \( e^{-0.25} \) using a calculator or a mathematical software: \[ e^{-0.25} \approx 0.7788 \] Thus, \( P(T > 500) \approx 0.7788 \). This is the probability that a single bulb lasts more than 500 hours.

Step 4: Define the binomial distribution for part b

Now, consider the case where at least 8 out of 10 bulbs last more than 500 hours. This is a binomial distribution problem with parameters \( n = 10 \) and \( p = P(T > 500) \approx 0.7788 \). We need to find: \[ P(X \ge 8) = P(X = 8) + P(X = 9) + P(X = 10) \]

Step 5: Calculate individual binomial probabilities

Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Calculate for \( k = 8 \), \( k = 9 \), and \( k = 10 \).

Step 6: Evaluation of binomial terms

For \( k = 8 \): \[ P(X = 8) = \binom{10}{8} (0.7788)^8 (0.2212)^2 \approx 0.233 \] For \( k = 9 \): \[ P(X = 9) = \binom{10}{9} (0.7788)^9 (0.2212)^1 \approx 0.392 \] For \( k = 10 \): \[ P(X = 10) = \binom{10}{10} (0.7788)^{10} (0.2212)^0 \approx 0.228 \]

Step 7: Sum the binomial probabilities

Sum the probabilities calculated in the previous step: \[ P(X = 8) + P(X = 9) + P(X = 10) \approx 0.233 + 0.392 + 0.228 = 0.853 \]

Key Concepts

Lebensdaueranalyse

Lebensdaueranalyse (Lifetime Analysis) deals with studying the length of time an item functions before it fails. In our problem, we are analyzing the lifespan of light bulbs using an exponential distribution. The exponential distribution is commonly used in reliability engineering and survival analysis because it describes the time between events in a process where events occur continuously and independently at a constant average rate.
The exponential distribution is characterized by its rate parameter, \[ \lambda \]. For our light bulbs, \[ \lambda \ = 5 \times 10^{-4} \]. This value tells us how often failures are expected to happen.
A key aspect of lifetime analysis is understanding the survival function, which gives the probability that a device or organism survives beyond a certain time. For an exponential distribution, the survival function S(t) at time t is given by:
\[ S(t) = e^{- \lambda t} \]
In part (a) of our problem, we calculated the probability that a single light bulb lasts more than 500 hours using this survival function, resulting in \[ S(500) = e^{-0.25} \approx 0.7788 \]. This means there is about a 77.88% chance that a bulb lasts more than 500 hours.

Binomialverteilung

The Binomialverteilung (Binomial Distribution) is used to model the number of successes in a fixed number of trials, where each trial has the same probability of success. In our problem, a 'success' is defined as a light bulb lasting more than 500 hours. We determined the success probability in part (a), which is \[ P(T > 500) \approx 0.7788 \].
For part (b), we need to find the probability that at least 8 out of 10 bulbs last more than 500 hours. This scenario is modeled using a binomial distribution with parameters \ n = 10 \ and \ p \approx 0.7788 \. The probability mass function for the binomial distribution is:
\[ P(X = k) = \binom{n}{k} \ p^k \ (1-p)^{n-k} \]
Using this, we calculated the probabilities for exactly 8, 9, and 10 bulbs lasting more than 500 hours and summed these probabilities to find the probability that at least 8 bulbs last more than 500 hours. The result is approximately 0.853, or 85.3%.

Überlebensfunktion

The Überlebensfunktion (Survival Function) provides the probability that a subject will survive past a certain time. It is a crucial component in fields like reliability engineering and actuarial science.
For an exponential distribution, the survival function at time t is given by:
\[ S(t) = 1 - F(t) = e^{- \lambda t} \]
In our exercise, we use the survival function to determine the probability that a light bulb lasts over 500 hours. Given \ \lambda = 5 \times 10^{-4} \, we calculated:
\[ P(T > 500) = e^{- \lambda \ (500)} = e^{-0.25} \approx 0.7788 \]
This tells us that there's about a 77.88% chance that a single bulb will last beyond 500 hours. The survival function is pivotal in longevity studies because it helps us understand and predict the lifespan of various items under constant failure rate assumptions.