Problem 26 - Calculus Concepts

Question

Consider the density function $$ g(t)=\left\\{\begin{array}{ll} 0.25 & \text { when } 5 \leq t \leq 9 \\ 0 & \text { when } t<5 \text { or } t>9 \end{array}\right. $$ a. Write \(G\), the corresponding cumulative distribution function. b. Use both \(g\) and \(G\) to calculate the probability that \(t<6.08\) c. Sketch graphs of \(g\) and \(G\).

Step-by-Step Solution

Answer

The cumulative distribution function is \(G(t)\). The probability that \(t < 6.08\) is 0.27. Graphs are piecewise for both \(g(t)\) and \(G(t)\).

Step 1: Identify the Nature of g(t)

The given function \(g(t)\) is a uniform density function defined over the interval \([5, 9]\). Since it is a density function, the total area under the function over its domain must be equal to 1.

Step 2: Determine the Properties of G(t)

The cumulative distribution function \(G(t)\) is the integral of \(g(t)\). For \(t < 5\), \(G(t) = 0\). For \(5 \leq t \leq 9\), since \(g(t) = 0.25\), \(G(t) = \int_{5}^{t} 0.25 \, dt = 0.25(t-5)\). For \(t > 9\), \(G(t) = 1\).

Step 3: Calculate Probability for t < 6.08 Using g(t)

The probability that \(t\) is less than 6.08 using \(g(t)\) is the area under the density function from 5 to 6.08. Thus, \(P(t < 6.08) = 0.25 \times (6.08 - 5) = 0.25 \times 1.08 = 0.27\).

Step 4: Calculate Probability for t < 6.08 Using G(t)

Using \(G(t)\), we find \(G(6.08) = 0.25(6.08 - 5) = 0.27\), which gives the cumulative probability up to 6.08.

Step 5: Sketch the Graph of g(t)

The graph of \(g(t)\) is a horizontal line at \(y = 0.25\) between \(x = 5\) and \(x = 9\). The line is 0 for \(x < 5\) and \(x > 9\).

Step 6: Sketch the Graph of G(t)

The graph of \(G(t)\) is a piecewise linear function: 0 for \(x < 5\), increasing linearly from 5 to 9 (beginning at 0 and ending at 1), and constant at 1 for \(x > 9\).

Key Concepts

density function

A density function, in the context of probability and statistics, represents how probabilities are distributed across a range of values. For a continuous random variable, the density function is often denoted by a lowercase letter, such as \( g(t) \). This function must satisfy the condition where the total area under the curve over its range equals 1. This signifies the total probability, ensuring we are capturing all possible outcomes.

In the provided exercise, the density function \( g(t) \) is defined within a specific interval from 5 to 9, where it maintains a constant value of 0.25. Outside this interval, the density function is 0, signifying that probabilities only exist within the set boundaries. Such a function is often referred to as a uniform density function in this specific range.

The constant value of 0.25 implies that the probability is evenly distributed over the interval. Calculating the area under this density function from 5 to 9 should equal 1, confirming the integral of the density function over this interval results in total probability. To check, compute the area using the formula for the area of a rectangle: height (0.25) times the width (9-5), which equals 1. This supports the principle that the total probability must sum to one over the defined interval.

uniform distribution

Uniform distribution describes a scenario where all outcomes are equally likely over a certain range. This concept is straightforward yet crucial in understanding probability distributions. In the exercise, the random variable \( t \) is uniformly distributed between 5 and 9. This means, within this interval, each subinterval of the same length is equally likely to occur.

The uniform distribution is characterized by two parameters: the minimum value (5 in this case) and the maximum value (9). The probability density function (PDF) for a uniform distribution is constant, reflecting the equal likelihood of all possible values within the specified interval.

The graph of a uniformly distributed density function is a rectangle over the range, flat at the constant density value. In this exercise, the uniform distribution allows for straightforward calculations, such as determining probabilities for various scenarios because the density function remains constant across the interval. This constancy simplifies the integration process when finding the cumulative distribution function (CDF), as demonstrated in this problem.

Using the example, the calculation of probability involved multiplying the density (0.25) by the range of interest, confirming uniform distributions' fundamental nature where probability calculations are simplified within the defined limits.

probability calculation

Probability calculation involves determining the likelihood of a random variable falling within a certain range. In this exercise, calculations were done using both the density function \( g(t) \) and the cumulative distribution function (CDF) \( G(t) \).

For values within the specified interval \([5, 9]\), the probability that \( t \) takes on a value less than 6.08 can be calculated in two ways. Using the density function, we calculate the area under \( g(t) \) from 5 to 6.08. This is done by multiplying the constant density (0.25) by the range of interest (6.08 - 5). This process provides a probability of 0.27.

Using the cumulative distribution function is another method. The CDF, \( G(t) \), is the integral of the density function, giving us the cumulative probability up to any point within the interval. In this case, \( G(6.08) = 0.25(6.08 - 5) \), also yielding a probability of 0.27.

Both methods illustrate how, depending on the situation, either function can be used effectively for probability calculations. The CDF presents an accumulated view of probabilities, advantageous for quickly assessing probabilities up to a particular point.