In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random variable X having a gar distribution with mean μ = 18 and variance o² = 108. (a) Find the values of a and B. (b) Find the probability that on any given day the daily power consumption will exceed 11 million kilowatt-hours. (a) α = B= (b) The probability is (Round to four decimal places as needed.)
In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random variable X having a gar distribution with mean μ = 18 and variance o² = 108. (a) Find the values of a and B. (b) Find the probability that on any given day the daily power consumption will exceed 11 million kilowatt-hours. (a) α = B= (b) The probability is (Round to four decimal places as needed.)
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
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![### Topic: Gamma Distribution in Daily Electric Power Consumption
#### Problem Statement
In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random variable \(X\) having a gamma distribution with mean \(\mu = 18\) and variance \(\sigma^2 = 108\).
(a) Find the values of \(\alpha\) and \(\beta\).
(b) Find the probability that on any given day the daily power consumption will exceed 11 million kilowatt-hours.
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#### Solution
##### (a) Determining \(\alpha\) and \(\beta\):
- **Formula for the Mean (\(\mu\))**:
\[ \mu = \alpha \beta \]
- **Formula for the Variance (\(\sigma^2\))**:
\[ \sigma^2 = \alpha \beta^2 \]
Given:
\[ \mu = 18 \]
\[ \sigma^2 = 108 \]
Using the formulas:
\[ 18 = \alpha \beta \]
\[ 108 = \alpha \beta^2 \]
- Solving for \(\alpha\) and \(\beta\):
1. From the mean formula,
\[ \alpha = \frac{18}{\beta} \]
2. Substitute \(\alpha\) in the variance formula:
\[ 108 = \left(\frac{18}{\beta}\right) \beta^2 \]
\[ 108 = 18 \beta \]
\[ \beta = 6 \]
3. Substitute \(\beta = 6\) back in to find \(\alpha\):
\[ \alpha = \frac{18}{6} = 3 \]
Therefore:
\[ \alpha = 3 \]
\[ \beta = 6 \]
##### (b) Calculating the Probability:
To find the probability that the daily power consumption exceeds 11 million kilowatt-hours:
\[ P(X > 11) \]
Since \(X\) follows a gamma distribution with parameters \(\alpha = 3\) and \(\beta = 6\):
Use the cumulative distribution function (CDF) of the gamma distribution or look up the value in gamma distribution tables. Alternatively, using statistical software or a calculator:
\[ P(X > 11) = 1 - P(X \leq 11) \]
Use the gamma CDF to find \( P](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d1e5150-1781-42a0-9e07-3d8cfcc0113b%2Ff181fb5f-7d32-4ec5-b29f-997cc20a145f%2Fchqsxif_processed.png&w=3840&q=75)
Transcribed Image Text:### Topic: Gamma Distribution in Daily Electric Power Consumption
#### Problem Statement
In a certain city, the daily consumption of electric power, in millions of kilowatt-hours, is a random variable \(X\) having a gamma distribution with mean \(\mu = 18\) and variance \(\sigma^2 = 108\).
(a) Find the values of \(\alpha\) and \(\beta\).
(b) Find the probability that on any given day the daily power consumption will exceed 11 million kilowatt-hours.
---
#### Solution
##### (a) Determining \(\alpha\) and \(\beta\):
- **Formula for the Mean (\(\mu\))**:
\[ \mu = \alpha \beta \]
- **Formula for the Variance (\(\sigma^2\))**:
\[ \sigma^2 = \alpha \beta^2 \]
Given:
\[ \mu = 18 \]
\[ \sigma^2 = 108 \]
Using the formulas:
\[ 18 = \alpha \beta \]
\[ 108 = \alpha \beta^2 \]
- Solving for \(\alpha\) and \(\beta\):
1. From the mean formula,
\[ \alpha = \frac{18}{\beta} \]
2. Substitute \(\alpha\) in the variance formula:
\[ 108 = \left(\frac{18}{\beta}\right) \beta^2 \]
\[ 108 = 18 \beta \]
\[ \beta = 6 \]
3. Substitute \(\beta = 6\) back in to find \(\alpha\):
\[ \alpha = \frac{18}{6} = 3 \]
Therefore:
\[ \alpha = 3 \]
\[ \beta = 6 \]
##### (b) Calculating the Probability:
To find the probability that the daily power consumption exceeds 11 million kilowatt-hours:
\[ P(X > 11) \]
Since \(X\) follows a gamma distribution with parameters \(\alpha = 3\) and \(\beta = 6\):
Use the cumulative distribution function (CDF) of the gamma distribution or look up the value in gamma distribution tables. Alternatively, using statistical software or a calculator:
\[ P(X > 11) = 1 - P(X \leq 11) \]
Use the gamma CDF to find \( P
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