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When working with probability calculations in the context of an exponential distribution, it is crucial to understand the cumulative distribution function (CDF). This function helps determine the probability that a random variable is less than or equal to a specific value, an event often expressed as "at most." In the given problem, we explored the probabilities of a distance being at most 100 meters, at most 200 meters, and between 100 and 200 meters.

For exponential distributions, the CDF is given by the formula:

• \( F(x) = 1 - e^{-\lambda x} \)

Using this, we calculated:- The CDF at 100 meters gives us \( P(X \leq 100) = 1 - e^{-1.386} \approx 0.75 \)- The CDF at 200 meters results in \( P(X \leq 200) = 1 - e^{-2.772} \approx 0.91 \)To find the probability between 100 and 200 meters, we subtract these values: \( P(100 < X \leq 200) = 0.91 - 0.75 = 0.16 \). This calculation provides a clear picture of how such distances are distributed probabilistically.

The median of an exponential distribution represents the point at which the probability of the random variable being less than or equal to it is 0.5. This characteristic is significant because it divides the data into two equal halves. For the exponential distribution, the median is determined by determining the value of \( M \) for which

• \( F(M) = 0.5 \)

Rearranging the cumulative distribution function, we get:

• \( 1 - e^{-\lambda M} = 0.5 \)

This leads us to the equation:

• \( e^{-\lambda M} = 0.5 \)

Solving for \( M \), we use:

• \( -\lambda M = \ln(0.5) \)
• \( M = -\frac{\ln(0.5)}{\lambda} \)

In the context of our problem, with \( \lambda = 0.01386 \), the calculation gives us a median distance of approximately 50 meters. This means that there’s an equal probability (50%) that the kangaroo rat moves less than or more than 50 meters to its first territorial vacancy.

In exponential distributions, both the mean and the standard deviation are crucial statistical measures, often represented with the same value. This occurs because the standard deviation of an exponential distribution is equal to its mean:

• \( \sigma = \mu = \frac{1}{\lambda} \)

This equivalence simplifies calculations and aids in understanding variability within the distribution. In this exercise, \( \lambda \) was given as 0.01386, leading to both the mean and the standard deviation being 72.1 meters.

To calculate the probability that the distance exceeds the mean by more than two standard deviations, we focus on the survival function, which helps find the probability of being greater than a certain value:

• \( P(X > \mu + 2\sigma) \)

Substituting the values:

• \( \mu + 2\sigma = 72.1 + 2 \times 72.1 = 216.3 \).

Thus, using the survival function,

• \( P(X > 216.3) = e^{-\lambda \times 216.3} \approx 0.05 \)

This means there’s a 5% chance the kangaroo rat travels more than 216.3 meters.

The probability density function (PDF) is central to understanding how probability is distributed in a continuous random variable, such as those governed by the exponential distribution. The form of the PDF for exponential distribution is

• \( f(x) = \lambda e^{-\lambda x} \)

This function defines the likelihood of the random variable assuming a particular value. For the exponential function, it's crucial to recognize:

• The parameter \( \lambda \) determines the rate of decay
• The PDF is defined only for \( x \geq 0 \)

For example, in our scenario, the PDF shows how the probability density decreases as the distance increases from the birth site. Thus larger distances are less probable for immediate territorial vacancies. Understanding the PDF in the exponential context helps predict and analyze behavior over continuous intervals, such as knowing the focus on shorter distances for movement when territorial vacancies are usually nearby.