Problem 14 - Algebra 2

Question

For Exercises \(9-14\) , use the following information. An exponential distribution has a mean of \(0.5 .\) Find each probability. $$ x<2.5 $$

Step-by-Step Solution

Answer

The probability is approximately 0.9933.

Step 1: Understanding the Exponential Distribution

The exponential distribution is defined by its probability density function (PDF) as \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \), where \( \lambda \) is the rate parameter. The mean of an exponential distribution is \( \frac{1}{\lambda} \). For this exercise, the mean is given as 0.5.

Step 2: Calculate the Rate Parameter \( \lambda \)

Using the formula for the mean of the exponential distribution \( \frac{1}{\lambda} = 0.5 \), solve for \( \lambda \):\[ \lambda = \frac{1}{0.5} = 2 \]

Step 3: Find the Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) of an exponential distribution is given by \( F(x; \lambda) = 1 - e^{-\lambda x} \), which represents the probability that a random variable \( X \) is less than or equal to \( x \).

Step 4: Substitute Values into the CDF

We need to find \( P(X < 2.5) \), which is \( F(2.5; 2) \). Substitute \( x = 2.5 \) and \( \lambda = 2 \) into the CDF:\[ F(2.5; 2) = 1 - e^{-2 \times 2.5} = 1 - e^{-5} \]

Step 5: Calculate the Exponential Term

Calculate \( e^{-5} \):Use a calculator to find the value:\[ e^{-5} \approx 0.006737946999085467 \]

Step 6: Find the Probability

Substitute \( e^{-5} \) back into the CDF formula to find the probability:\[ P(X < 2.5) = 1 - e^{-5} \approx 1 - 0.006737946999085467 = 0.9932620530009145 \]

Step 7: Approximate to the Desired Precision

Round the probability to a reasonable level of precision, if necessary. Here, we can express the probability as:\[ P(X < 2.5) \approx 0.9933 \]

Key Concepts

Probability Density Function

The probability density function (PDF) is a fundamental concept in understanding distributions, especially for continuous data, like the exponential distribution. The PDF of the exponential distribution is defined as \( f(x; \lambda) = \lambda e^{-\lambda x} \), where \( x \geq 0 \) and \( \lambda \) is the rate parameter. This function describes the likelihood that a random variable \( X \) will fall within a particular range of values.

For the exponential distribution, the PDF has a few key characteristics:

• It is always non-negative, meaning \( f(x; \lambda) \geq 0 \) for all \( x \).
• The area under the curve is 1, which represents the total probability of all possible outcomes.
• The function decreases exponentially, indicating that as \( x \) increases, the probability of observing that value decreases.

In the context of this exercise, the rate parameter \( \lambda \) is 2, making the PDF of the exponential distribution \( f(x; 2) = 2e^{-2x} \). This means the likelihood decreases rapidly as \( x \) becomes larger.

Cumulative Distribution Function

The cumulative distribution function (CDF) helps us understand the probability that a random variable is less than or equal to a particular value. Unlike the PDF, which gives us a sense of density, the CDF is concerned with accumulation. For an exponential distribution, the CDF is given by \( F(x; \lambda) = 1 - e^{-\lambda x} \).

In simple terms, the CDF shows the probability of \( X \leq x \). This means it tells us the probability that the outcome will fall within a certain "cumulative" range up to \( x \).

• The CDF starts at 0 and increases towards 1 as \( x \) moves towards infinity.
• At \( x = 0 \), the CDF is \( F(0; \lambda) = 0 \), and as \( x \to \infty \), the CDF approaches 1.

In our problem, to find the probability \( P(X < 2.5) \), we compute the CDF value at 2.5 when \( \lambda = 2 \), resulting in \( F(2.5; 2) = 1 - e^{-5} \). This gives us a probability close to 0.9933.

Rate Parameter

The rate parameter, denoted \( \lambda \), is crucial in defining the characteristics of the exponential distribution. This parameter determines the spread or rate at which probabilities decrease relative to increasing \( x \) values.

The reciprocal of the rate parameter gives us the mean of the distribution. Therefore, \( \lambda \) is inversely related to the average amount of time until an event happens. In mathematical terms, the mean \( \mu \) is given as \( \frac{1}{\lambda} \).

• A higher \( \lambda \) results in a quicker drop of probabilities, meaning events are more likely to occur in shorter time spans.
• A lower \( \lambda \) makes the distribution flatter, indicating more spread out events over time.

In this exercise, with the mean provided as 0.5, we calculated \( \lambda = 2 \). This tells us that on average, events happen twice as fast as in a distribution with a rate of 1.