Question 1162676: 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! For an exponentially distributed random variable $X$, we use the properties of the exponential distribution where the mean is $\mu = \frac{1}{\lambda}$.
### a-1. What is the rate parameter $\lambda$?
The relationship between the expected value and the rate parameter is $E[X] = \frac{1}{\lambda}$.
$$\lambda = \frac{1}{67} \approx 0.014925$$
Rounding to 3 decimal places:
**$\lambda = 0.015$**
### a-2. What is the standard deviation of $X$?
For the exponential distribution, the standard deviation ($\sigma$) is equal to the mean ($E[X]$).
$$\sigma = 67$$
**Standard deviation = 67**
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### b. Compute $P(58 \leq X \leq 76)$
The cumulative distribution function (CDF) for an exponential distribution is $F(x) = 1 - e^{-\lambda x}$. To find the probability between two values, we use $P(a \leq X \leq b) = e^{-\lambda a} - e^{-\lambda b}$.
Using the precise $\lambda = \frac{1}{67} \approx 0.014925$:
* $e^{-(0.014925 \times 58)} = e^{-0.865671} \approx 0.4208$
* $e^{-(0.014925 \times 76)} = e^{-1.134328} \approx 0.3216$
$$P(58 \leq X \leq 76) = 0.42077 - 0.32164 = 0.09913$$
Rounding to 4 decimal places:
**$P(58 \leq X \leq 76) = 0.0991$**
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### c. Compute $P(55 \leq X \leq 79)$
Using the same method:
* $e^{-(0.014925 \times 55)} = e^{-0.820895} \approx 0.4401$
* $e^{-(0.014925 \times 79)} = e^{-1.179104} \approx 0.3076$
$$P(55 \leq X \leq 79) = 0.44005 - 0.30757 = 0.13248$$
Rounding to 4 decimal places:
**$P(55 \leq X \leq 79) = 0.1325$**
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