The time till failure for an exponentially distributed random variable can be modeled by the probability density function (PDF):

$f(t) = \lambda e^{-\lambda t}$

where $\lambda$ is the rate parameter, and $t$ represents time. The cumulative distribution function (CDF) for the exponential distribution, which gives the probability that the time till failure is less than or equal to some value $t$, is:

$F(t) = 1 - e^{-\lambda t}$

Given the problem:

• 12% of TVs fail by 1,600 hours, which translates to $F(1600) = 0.12$.
• We need to find the probability that TVs will still be working after 7,100 hours, i.e., $P(T > 7100)$.

Step 1: Determine the rate parameter ($\lambda$)

We know from the CDF that:

$F(1600) = 1 - e^{-\lambda \cdot 1600}$

Given $F(1600) = 0.12$, we can solve for $\lambda$:

$0.12 = 1 - e^{-\lambda \cdot 1600}$ $e^{-\lambda \cdot 1600} = 0.88$ $-\lambda \cdot 1600 = \ln(0.88)$ $\lambda = -\frac{\ln(0.88)}{1600}$

Step 2: Calculate the probability that the TV lasts longer than 7,100 hours

The probability that the TV will still be working after 7,100 hours is given by:

$P(T > 7100) = 1 - F(7100) = e^{-\lambda \cdot 7100}$

Let's calculate $\lambda$ and the final probability. I'll do the math now.The rate parameter ($\lambda$) for the exponential distribution is approximately $7.99 \times 10^{-5}$.

The probability that a TV will still be working after 7,100 hours is approximately 0.567, or 56.7%.

Would you like more details on any step or further clarifications?

Here are 5 related questions you might find interesting:

1. How is the exponential distribution applied in other real-world failure models?
2. What is the expected time until failure based on the calculated $\lambda$?
3. How do other distributions, like the Weibull distribution, compare to the exponential distribution in modeling failure?
4. Can we calculate the probability that a TV fails between two given time intervals?
5. How does increasing the rate parameter $\lambda$ affect the lifespan of devices?

Tip: In exponential distributions, the expected value (mean) is the inverse of the rate parameter $\lambda$, which gives a quick estimate of the average time to failure.