SOLUTION: A random variable X is exponentially distributed with an expected value of 67. a-1. What is the rate parameter λ? (Round your answer to 3 decimal places.) a-2. What i

Question 1162932: A random variable X is exponentially distributed with an expected value of 67.

a-1. What is the rate parameter λ? (Round your answer to 3 decimal places.)

a-2. What is the standard deviation of X?

b. Compute P(58 ≤ X ≤ 76). (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

c. Compute P(55 ≤ X ≤ 79). (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

Answer by CPhill(2261) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import math
# Given
mean_x = 67
# a-1. Rate parameter lambda
lambda_val = 1 / mean_x
lambda_rounded = round(lambda_val, 3)
# a-2. Standard deviation
std_dev = mean_x
# b. P(58 <= X <= 76)
# P(a <= X <= b) = F(b) - F(a) = (1 - exp(-lambda*b)) - (1 - exp(-lambda*a)) = exp(-lambda*a) - exp(-lambda*b)
a_b = 58
b_b = 76
prob_b = math.exp(-lambda_val * a_b) - math.exp(-lambda_val * b_b)
# c. P(55 <= X <= 79)
a_c = 55
b_c = 79
prob_c = math.exp(-lambda_val * a_c) - math.exp(-lambda_val * b_c)
print(f"{lambda_val=}")
print(f"{lambda_rounded=}")
print(f"{std_dev=}")
print(f"{prob_b=}")
print(f"{prob_c=}")

```
```text?code_stdout&code_event_index=2
lambda_val=0.014925373134328358
lambda_rounded=0.015
std_dev=67
prob_b=0.09913077737880754
prob_c=0.132483377810241
```
For an exponentially distributed random variable $X$:
### a-1. Rate parameter $\lambda$
The relationship between the expected value (mean) and the rate parameter $\lambda$ is $E[X] = \frac{1}{\lambda}$.
$$\lambda = \frac{1}{67} \approx 0.014925$$
Rounding to 3 decimal places:
**$\lambda = 0.015$**
### a-2. Standard deviation of $X$
For an exponential distribution, the standard deviation is equal to the mean.
$$\sigma = E[X] = 67$$
**Standard deviation = 67**
---
### b. Compute $P(58 \leq X \leq 76)$
The cumulative distribution function (CDF) for an exponential distribution is $F(x) = 1 - e^{-\lambda x}$.
The probability $P(a \leq X \leq b)$ is calculated as $F(b) - F(a) = e^{-\lambda a} - e^{-\lambda b}$.
Using $\lambda = \frac{1}{67} \approx 0.014925$:
* $e^{-(0.014925 \times 58)} \approx e^{-0.8657} \approx 0.4208$
* $e^{-(0.014925 \times 76)} \approx e^{-1.1343} \approx 0.3216$
$$P(58 \leq X \leq 76) = 0.4208 - 0.3216 = 0.0992$$
*(Using more precise intermediate values: $0.42077 - 0.32164 = 0.09913$)*
Rounding to 4 decimal places:
**$P(58 \leq X \leq 76) = 0.0991$**
---
### c. Compute $P(55 \leq X \leq 79)$
Using the same formula: $P(55 \leq X \leq 79) = e^{-\lambda(55)} - e^{-\lambda(79)}$
* $e^{-(0.014925 \times 55)} \approx e^{-0.8209} \approx 0.4400$
* $e^{-(0.014925 \times 79)} \approx e^{-1.1791} \approx 0.3076$
$$P(55 \leq X \leq 79) = 0.4400 - 0.3076 = 0.1324$$
*(Using more precise intermediate values: $0.44005 - 0.30757 = 0.13248$)*
Rounding to 4 decimal places:
**$P(55 \leq X \leq 79) = 0.1325$**
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