SOLUTION: A professor sees students during regular office hours. Time spent with students follows an exponential distribution with a mean of 10 minutes.
a. Find the probability that a given
Algebra ->
Distributive-associative-commutative-properties
-> SOLUTION: A professor sees students during regular office hours. Time spent with students follows an exponential distribution with a mean of 10 minutes.
a. Find the probability that a given
Log On
Question 1160844: A professor sees students during regular office hours. Time spent with students follows an exponential distribution with a mean of 10 minutes.
a. Find the probability that a given student spends fewer than 20 minutes with the professor.
b. Find the probability that a given student spends more than 5 minutes with the professor.
c. Find the probability that a given student spends between 10 and 15 minutes with the professor. Answer by CPhill(2303) (Show Source):
You can put this solution on YOUR website! ```python
import math
lam = 0.1
# a. P(X < 20)
p_a = 1 - math.exp(-lam * 20)
# b. P(X > 5)
p_b = math.exp(-lam * 5)
# c. P(10 < X < 15)
p_c = math.exp(-lam * 10) - math.exp(-lam * 15)
print(f"a: 1 - e^(-2) = {p_a:.6f}")
print(f"b: e^(-0.5) = {p_b:.6f}")
print(f"c: e^(-1) - e^(-1.5) = {p_c:.6f}")
```
Let $X$ denote the time (in minutes) a student spends with the professor.
Since $X$ follows an exponential distribution with a mean of $\mu = 10\text{ minutes}$, the rate parameter is:
$$\lambda = \frac{1}{\mu} = \frac{1}{10} = 0.1\text{ per minute}$$
For an exponential random variable $X \sim \text{Exp}(\lambda)$:
* **Cumulative Distribution Function (CDF):** $P(X \le x) = 1 - e^{-\lambda x}$
* **Survival Function:** $P(X > x) = e^{-\lambda x}$
---
### a. Probability that a student spends fewer than 20 minutes
$$P(X < 20) = 1 - e^{-0.1 \times 20} = 1 - e^{-2}$$
$$P(X < 20) \approx 1 - 0.135335 = \mathbf{0.8647} \quad (\approx 86.47\%)$$
---
### b. Probability that a student spends more than 5 minutes
$$P(X > 5) = e^{-0.1 \times 5} = e^{-0.5}$$
$$P(X > 5) \approx \mathbf{0.6065} \quad (\approx 60.65\%)$$
---
### c. Probability that a student spends between 10 and 15 minutes
$$P(10 < X < 15) = P(X \le 15) - P(X \le 10)$$
$$P(10 < X < 15) = (1 - e^{-0.1 \times 15}) - (1 - e^{-0.1 \times 10}) = e^{-1} - e^{-1.5}$$
$$P(10 < X < 15) \approx 0.367879 - 0.223130 = \mathbf{0.1447} \quad (\approx 14.47\%)$$
