Problem 11 - Algebra 2

Question

For Exercises \(9-14\) , use the following information. An exponential distribution has a mean of \(0.5 .\) Find each probability. $$ x>\frac{1}{4} $$

Step-by-Step Solution

Answer

The probability is approximately 0.6065.

Step 1: Explain the Exponential Distribution

An exponential distribution is defined by its mean \( \mu \), and its probability density function is given by \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \). The parameter \( \lambda \) is the rate parameter and is the reciprocal of the mean, \( \lambda = \frac{1}{\mu} \).

Step 2: Identify the Rate Parameter

Given that the mean \( \mu \) is 0.5, we can find the rate parameter \( \lambda \) of the exponential distribution:\[ \lambda = \frac{1}{0.5} = 2. \]

Step 3: Set Up the Probability Expression

We need to find the probability that \( X > \frac{1}{4} \). For an exponential distribution, the cumulative distribution function (CDF) is \( F(x; \lambda) = 1 - e^{-\lambda x} \). The probability that \( X > x \) is given by the survival function, which is the complement of the CDF:\[ P(X > x) = 1 - F(x; \lambda) = e^{-\lambda x}. \]

Step 4: Substitute and Calculate the Probability

Using the survival function and substituting \( \lambda = 2 \) and \( x = \frac{1}{4} \), we have:\[ P\left(X > \frac{1}{4}\right) = e^{-2 \times \frac{1}{4}} = e^{-0.5}. \]We can calculate this probability using \[ e^{-0.5} \approx 0.6065. \]

Step 5: Conclusion

Therefore, the probability that \( X > \frac{1}{4} \) in this exponential distribution is approximately 0.6065.

Key Concepts

Probability Density Function

The Probability Density Function (PDF) is a cornerstone concept when dealing with continuous random variables, such as those described by the exponential distribution. In the context of exponential distribution, the PDF is expressed as \[ f(x; \lambda) = \lambda e^{-\lambda x} \]where:

• \( x \) is a non-negative random variable, meaning \( x \geq 0 \).
• \( \lambda \) is the rate parameter. It controls the spread of the distribution. Higher values of \( \lambda \) indicate that events occur more frequently.

The PDF tells us how the probability of the random variable is distributed along its range of values. It does not give a probability at a single point but rather describes where outcomes are more or less likely to occur. To find actual probabilities within a specific range, integrating the PDF over that range is necessary. This mathematical integration provides the area under the curve, representing probabilities.

Mean and Rate Parameter

In the exponential distribution, the mean and the rate parameter have a special relationship. The mean, denoted as \( \mu \), represents the expected average time between events. It's a central tendency measure explaining how long we expect to wait between occurrences.
The rate parameter, \( \lambda \), is pivotal because it defines the frequency of these events and is reciprocal to the mean:\[ \lambda = \frac{1}{\mu} \]

• If the mean \( \mu \) is large, implying a longer wait, the rate \( \lambda \) becomes smaller.
• Conversely, a small mean results in a larger rate, indicating frequent events.

In simpler terms, \( \lambda \) tells us how often we can expect the occurrence of the observed event. For instance, if the mean is 0.5, then according to the formula, \( \lambda \) equals 2, denoting that on average, we expect two events per unit time.

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) provides a way to understand the probability of the random variable, \( X \), being less than or equal to a certain value. For an exponential distribution with a rate parameter \( \lambda \), the CDF is expressed as:\[ F(x; \lambda) = 1 - e^{-\lambda x} \]

• It quantifies the likelihood that \( X \) is less than or exactly \( x \).
• The function increases as \( x \) increases, approaching 1 since it represents probabilities that always accumulate.

For reverse inquiries, such as finding \( P(X > x) \), which is referred to as the survival function, this can also be denoted as:\[ P(X > x) = e^{-\lambda x} \]This expression calculates the probability of the variable being greater than a given number \( x \), effectively making use of the complement rule \( 1 - F(x) \). In our specific problem, to find \( P(X > \frac{1}{4}) \), we substitute in our formula to determine the exact probability, which describes how much beyond \( \frac{1}{4} \) the variable is expected to lie.