We are given that the time till failure of a particular TV brand follows an exponential distribution. The exponential distribution is described by the probability density function:

$f(t) = \lambda e^{-\lambda t}, \quad t \geq 0$

Where:

• $\lambda$ is the rate parameter of the exponential distribution.
• The cumulative distribution function (CDF) gives the probability that the TV will fail by time $t$:

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

Given Information:

• 12% of TVs have failed by 1,600 hours, which gives us:

$P(T \leq 1600) = 0.12$

This can be written using the CDF as:

$1 - e^{-\lambda \cdot 1600} = 0.12$

We will solve this equation to find $\lambda$, and then use it to calculate the probability that the TV will still be working after 7,100 hours.

Step 1: Solve for $\lambda$

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

Step 2: Find the probability that the TV will still be working after 7,100 hours

The probability that the TV will still be working after 7,100 hours is the complement of the probability that it fails by 7,100 hours:

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

Let's calculate these values 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 further details or clarifications? Here are some related questions for deeper understanding:

1. How does the exponential distribution describe the time until an event occurs?
2. How would the probability change if 50% of TVs failed by 1600 hours instead of 12%?
3. What are the implications of a higher or lower rate parameter $\lambda$ in an exponential distribution?
4. How can we calculate the expected time till failure for this TV brand?
5. What are the limitations of using the exponential distribution to model TV failures?

Tip: In the exponential distribution, the mean time to failure is given by $\frac{1}{\lambda}$, which is a useful measure of expected lifetime.