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Suppose the random variable \(x\) has an exponential probability distribution with \(\theta=3\). Find the mean and standard deviation of \(x\). Find the probability that \(x\) will assume a value within the interval \(\mu \pm 2 \sigma\).

Short Answer

Expert verified
The mean and standard deviation are both 3. The probability of \(x\) being within \([0, 9]\) is approximately 95%.

Step by step solution

01

Understanding the Exponential Distribution

The exponential distribution is defined for a random variable \(x\) with the parameter \(\theta\). It has the probability density function \( f(x; \theta) = \frac{1}{\theta} e^{-x/\theta} \) for \( x \geq 0 \). In this exercise, \( \theta \) is given as 3.
02

Calculating the Mean

The mean of an exponential distribution is given by \( \mu = \theta \). For \( \theta = 3 \), the mean \( \mu = 3 \).
03

Calculating the Standard Deviation

The standard deviation of an exponential distribution is the same as its mean, thus \( \sigma = \theta \). For \( \theta = 3 \), the standard deviation \( \sigma = 3 \).
04

Defining the Interval \( \mu \pm 2\sigma \)

The interval \( \mu \pm 2\sigma \) means \( 3 \pm 2 \times 3 \). This simplifies to an interval of \([-3, 9]\). However, since \(x\) in an exponential distribution cannot be negative, we adjust the interval to \([0, 9]\).
05

Calculating the Probability within the Interval

The cumulative distribution function (CDF) for the exponential distribution is \( F(x; \theta) = 1 - e^{-x/\theta} \). To find \( P(0 \leq x \leq 9) \), calculate this as follows: \( P(x \leq 9) = 1 - e^{-9/3} = 1 - e^{-3} \). For \( x = 0 \), \( P(x \leq 0) = 1 - e^{0} = 0 \). Thus, \( P(0 \leq x \leq 9) = 1 - e^{-3} \).
06

Evaluating the Cumulative Probability

The numerical value of \( e^{-3} \) is approximately 0.0498. Thus, \( P(0 \leq x \leq 9) = 1 - 0.0498 = 0.9502 \), or approximately 95%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distributions
Probability distributions are essential in statistics as they describe how the probabilities are spread out over the possible values of a random variable. A good understanding of probability distributions helps predict outcomes and make informed decisions.

In the context of exponential distribution, which is often used to model the time until an event occurs, the distribution is defined for a positive random variable \(x\). It is characterized by a single parameter \(\theta\), often referred to as the rate parameter. This distribution is crucial when modeling processes that occur continuously and independently at a constant average rate.

An exponential distribution's probability density function (PDF) is given by:
  • \( f(x; \theta) = \frac{1}{\theta} e^{-x/\theta} \) for \( x \geq 0 \)
These components help us understand how likely different outcomes are, facilitating predictions regarding the time or location of future events.
Mean and Standard Deviation
The mean and standard deviation are vital concepts when exploring any distribution, including exponential distribution. They provide insights into the distribution's center and variability, respectively.

For an exponential distribution, the mean \( \mu \) and the standard deviation \( \sigma \) are both equal to the parameter \( \theta \). Therefore, if \( \theta = 3 \), as mentioned in our exercise, we calculate:
  • Mean (\( \mu \)) \( = \theta = 3\)
  • Standard deviation (\( \sigma \)) \( = \theta = 3\)
Both statistics indicate that the data points tend to cluster around the mean of 3 and spread out to the same degree as this mean.

Using these values, you can define intervals such as \( \mu \pm 2\sigma \), helping assess the range where a majority of values lie, effectively enhancing the understanding of variability in the distribution.
Cumulative Distribution Function
The cumulative distribution function (CDF) describes the probability that a random variable \( x \) will take a value less than or equal to \( x \). It is a crucial tool in probability for assessing cumulative probability. For an exponential distribution, the CDF is expressed as:
  • \( F(x; \theta) = 1 - e^{-x/\theta} \)
This function helps determine probabilities over intervals. For instance, considering the exercise, to find the probability that \( x \) falls within the interval \([0, 9]\), we use the CDF.

Calculate:
  • \( P(x \leq 9) = 1 - e^{-9/3} = 1 - e^{-3} \)
  • \( P(x \leq 0) = 1 - e^{0} = 0 \)
As you compute, \( 1 - e^{-3} \approx 0.9502 \), indicating a 95% probability. This CDF provides comprehensive insight into the probability landscape of a random variable over specific intervals, aiding in effective decision-making based on these probabilities.

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Most popular questions from this chapter

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