Problem 20 - Calculus Concepts | StudyQuestionHub

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Problem 20

Question

Verify the following statements for the uniform density function $$ u(x)=\left\\{\begin{array}{ll} \frac{1}{b-a} & \text { when } a \leq x \leq b \\ 0 & \text { when } xb \end{array}\right. $$ a. The mean is \(\mu=\frac{a+b}{2}\). b. The standard deviation is \(\sigma=\frac{b-a}{\sqrt{12}}\). c. The cumulative distribution function is $$ F(x)=\left\\{\begin{array}{ll} 0 & \text { when } x<0 \\ \frac{x-a}{b-a} & \text { when } a \leq x \leq b \\ 1 & \text { when } x>b \end{array}\right. $$

Step-by-Step Solution

All statements are correct based on the properties of uniform distribution.

1Step 1: Calculate the Mean

For a uniform distribution on the interval \([a, b]\), the mean \( \mu \) is given by the formula \( \mu = \frac{a+b}{2} \). This represents the average value of \(x\) over the interval, and it intuitively makes sense because it's the midpoint of the interval \([a, b]\).

2Step 2: Calculate the Variance

The variance \( \sigma^2 \) for a uniform distribution on \([a, b]\) is calculated using the formula \( \sigma^2 = \frac{(b-a)^2}{12} \). This formula derives from the nature of the distribution spread evenly across \([a, b]\).

3Step 3: Calculate the Standard Deviation

The standard deviation \( \sigma \) is simply the square root of the variance. Using the result from the previous step, \( \sigma = \sqrt{\frac{(b-a)^2}{12}} = \frac{b-a}{\sqrt{12}} \).

4Step 4: Derive the Cumulative Distribution Function (CDF)

To find the cumulative distribution function (CDF) for a uniform distribution on \([a, b]\), consider the following cases:1. For \( x < a \), the CDF \( F(x) = 0 \), since no values of \(x\) are captured.2. For \( a \leq x \leq b \), the CDF is the integral of the probability density function (PDF) from \( a \) to \( x \):\[ F(x) = \int_a^x \frac{1}{b-a} \, dt = \frac{x-a}{b-a} \]3. For \( x > b \), the CDF \( F(x) = 1 \), as all the values up to \( b \) are captured.

Key Concepts

Probability Density FunctionCumulative Distribution FunctionMeanStandard Deviation

Probability Density Function

The probability density function (PDF) of a uniform distribution describes how probabilities are distributed across the range of possible outcomes. For a uniform distribution defined on the interval \([a, b]\), the PDF is expressed as:

\(u(x) = \frac{1}{b-a}\) when \(a \leq x \leq b\)
\(u(x) = 0\) when \(x < a\) or \(x > b\)

This tells us that every value within the interval \([a, b]\) is equally likely. It results in a rectangular distribution where the height is constant over the interval, providing a uniform level of probability. Since the total area under the PDF must equal 1, the height of the rectangle is \(\frac{1}{b-a}\). It's important to remember that a PDF does not give probabilities of specific values but rather describes how probability density is distributed.

Cumulative Distribution Function

The cumulative distribution function (CDF) of a uniform distribution gives the probability that the random variable \(X\) is less than or equal to some value \(x\). This is represented by the following piece of wise function:

\(F(x) = 0\) for \(x < a\)
\(F(x) = \frac{x-a}{b-a}\) for \(a \leq x \leq b\)
\(F(x) = 1\) for \(x > b\)

These conditions intuitively capture the essence of increasing probability, as \(x\) expands from \(a\) to \(b\). For values of \(x\) below \(a\), no probability is accumulated, hence \(F(x) = 0\). As \(x\) progresses within the interval, the probability increases linearly, reaching \(1\) once \(x\) surpasses \(b\). The CDF effectively provides a comprehensive view of probability up to any point \(x\) along the interval.

Mean

The mean of a uniform distribution, often denoted as \(\mu\), is the average of all possible outcomes within the interval. For a uniform distribution across the range \([a, b]\), the formula for the mean is:\[\mu = \frac{a+b}{2}\]This is because the distribution is symmetric around its central point. Thus, the mean simply lies halfway between \(a\) and \(b\). It's a straightforward concept as the distribution is uniform and symmetric, leading to a perfectly centered average. Understanding the mean helps identify the 'central' location or 'center of mass' of the distribution.

Standard Deviation

Standard deviation is a measure of how spread out the values in a distribution are. In the context of a uniform distribution from \([a, b]\), the standard deviation \(\sigma\) can be calculated using the formula:\[\sigma = \frac{b-a}{\sqrt{12}}\]This formula comes from taking the square root of the variance, which measures the average squared distance of the distribution's values from the mean. The factor \(\sqrt{12}\) comes from the derivation process, capturing the uniform nature. A higher standard deviation suggests a wider spread of values, while a smaller one indicates a more central-tendency with values closer together. Understanding the standard deviation is critical for assessing the risk or variability in processes modeled by uniform distribution.

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