We are given the probability density function (pdf) $p(t) = 0.24 e^{-0.24t}$ to model the time $t$ taken to complete a task in minutes, with $t \in [0, 50]$.

Part A: Finding the Median

The median is the value $M$ such that the probability of the task being completed in $t \leq M$ minutes is 0.5. Mathematically, we solve: $\int_0^M p(t) \, dt = 0.5$ Substituting the pdf $p(t) = 0.24 e^{-0.24t}$, we need to compute: $\int_0^M 0.24 e^{-0.24t} \, dt = 0.5$

Part B: Finding the Mean

The mean or expected value $\mathbb{E}(T)$ is calculated as: $\mathbb{E}(T) = \int_0^{50} t \cdot p(t) \, dt = \int_0^{50} t \cdot 0.24 e^{-0.24t} \, dt$

I'll solve both parts to find the median and mean values.

Let me compute these integrals.### Results:

• Part A: Median number of minutes taken is approximately 2.89 minutes.
• Part B: Mean number of minutes taken is approximately 4.17 minutes.

Would you like a detailed explanation of the steps involved? Here are some related questions for further exploration:

1. How do we interpret the difference between the median and mean in this context?
2. What would happen to the mean if the upper limit of 50 minutes was extended to infinity?
3. How does the exponential decay factor $e^{-0.24t}$ influence the shape of the probability distribution?
4. Can we calculate the variance of this distribution to better understand its spread?
5. What is the cumulative distribution function (CDF) for this probability density function?

Tip: The median of a distribution is where half of the probability lies to the left, while the mean represents the "center of mass" of the probability distribution. For skewed distributions like the exponential, the median is often less than the mean.