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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.

The density function for the time to complete a certain task is approximately equal to p(t) = 0.24e^(-0.24t), where t is time in minutes and 0 ≤ t ≤ 50. A. Find the median number of minutes taken, to 2 decimal places. B. Find the mean number of minutes taken, to 2 decimal places.

Solve for the median and mean number of minutes taken using the probability density function p(t) = 0.24e^(-0.24t) for t in the range of 0 to 50 minutes. This problem involves key concepts like exponential distributions and integral calculus.

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