Given probubility density function F(t) = et tzopa oimebooA qoT Find t=L where typical device is 600% likely to exceed Student A : S et dt O.6= Student B: O.6 0.6 =1- e-t dt
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![**Given Probability Density Function**
\[ F(t) = e^{-t} \quad t \geq 0 \]
**Objective:**
Find \( t = L \) where a typical device is 60% likely to exceed.
**Student A's Approach:**
\[ 0.6 = \int_{L}^{\infty} e^{-t} \, dt \]
**Student B's Approach:**
\[ 0.6 = 1 - \int_{0}^{L} e^{-t} \, dt \]
The problem involves calculating the time \( L \) at which a device has a 60% probability of lasting beyond. Student A integrates over \( t \) from \( L \) to infinity, modeling the probability of survival beyond \( L \). Student B calculates the cumulative probability up to \( L \) and subtracts it from 1 to find the likelihood of exceeding \( L \). Both approaches aim to find the same value of \( L \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb4f8469-eda0-4bfd-b67e-0d596825d7c6%2Fdcc985e9-e9fa-4962-ad77-63faf784be7a%2F2rhgtjf_processed.jpeg&w=3840&q=75)
Consider the given information.
First calculate for the student A.
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