Radio has been an important source of entertainment and information for a long time. People rely on radios for various purposes. But, one of the biggest concerns for radio owners is its lifespan. A radio's lifetime is usually modeled as an exponential distribution with a mean of ten years. In this post, we will examine the probability that a ten-year-old radio will be working after an additional ten years.
Problem Statement:
The lifetime of a radio is exponentially distributed with a mean of ten years. If Jones buys a ten-year-old radio, what is the probability that it will be working after an additional ten years?
Solution:
The exponential distribution is a continuous distribution that is often used to model the time between events in a Poisson process. The exponential distribution is characterized by its rate parameter, μ, which is the reciprocal of the mean. In this case, the mean is ten years, so μ = 1/10.
The probability that a radio will be working after an additional ten years can be calculated using the cumulative distribution function of the exponential distribution:
F(x) = 1 - e^(-μx)
Where x is the time elapsed, and F(x) is the cumulative probability of the radio still being operational.
In this case, we want to find the probability that the radio will still be working after ten additional years, so x = 10.
F(10) = 1 - e^(-μ * 10)
= 1 - e^(-1/10 * 10)
= 1 - e^(-1)
= 1 - 0.368
= 0.632
So, the probability that Jones's radio will still be working after ten additional years is 0.632 or 63.2%.
We have seen how to use the exponential distribution to model the lifetime of a radio. Given a ten-year-old radio, we have found that the probability that it will be working after an additional ten years is 63.2%. This post provides a simple example of how the exponential distribution can be used to make predictions about the longevity of a product.
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