91Ó°ÊÓ

In the world of statistics, the Probability Density Function, or PDF, is an essential concept used to specify the probability of a random variable falling within a particular range of values. For continuous random variables like in our exercise, the PDF provides a curve that represents probabilities. It does not give probabilities directly but rather allows us to compute them over an interval by integrating the function.
In the case of an exponential distribution, the PDF is defined as:

• For any positive value of the random variable, say \( x \), the PDF is given by \( f(x) = \frac{1}{ heta} e^{-x/\theta} \), where \( \theta \) is the scale parameter, equivalent to the mean \( 1/\lambda \).
• For \( x \leq 0 \), the PDF value is 0 since the exponential distribution is not defined for negative values.

Therefore, in our problem, the PDF \( f_{X}(x) = \frac{1}{30} e^{-x/30} \) represents an exponential distribution with a rate parameter \( \lambda = \frac{1}{30} \). This parameter is crucial as it influences how rapidly the probabilities taper off as you move away from zero.

The Cumulative Distribution Function, or CDF, is a crucial tool in statistics to find the probability that a random variable \( X \) is less than or equal to a certain value, say \( x \). It's essentially the cumulative sum of probabilities up to \( x \).
For an exponential distribution, the CDF is expressed as:

• \( F(x) = 1 - e^{-\lambda x} \) for \( x > 0 \), where \( \lambda \) is the rate parameter.
• For \( x \leq 0 \), the CDF is 0, reflecting no probability of "negative" values for a positive process.

In our exercise, the calculation of \( P(X \leq 19) \) involves plugging 19 into the CDF: \( F(19) = 1 - e^{-19/30} \). Solving this gives us the probability that a tire lasts at most 19000 km, which is helpful for decision-making and planning.

The mean and standard deviation are fundamental characteristics of a distribution, offering insights into its central tendency and spread.
For the exponential distribution, both these values are described with a simple relationship to the rate parameter \( \lambda \):

• The mean \( \mu \) is given by \( \frac{1}{\lambda} \). This is the expected value, or the long-term average of the random variable. In our exercise, with \( \lambda = \frac{1}{30} \), the mean distance a tire lasts is 30 thousand km.
• Interestingly, the standard deviation \( \sigma \) is also \( \frac{1}{\lambda} \), which indicates that the spread or variability in the data is identical to the mean for exponential distributions.

This simplicity makes the exponential distribution unique and interesting, with the mean and standard deviation providing a quick snapshot of the distribution's behavior.

The Interquartile Range (IQR) is a measure of statistical dispersion, which indicates the spread of the middle 50% of a dataset. It is the difference between the 75th percentile (\( Q3 \)) and the 25th percentile (\( Q1 \)).
For an exponential distribution, finding these percentiles involves the CDF:

• \( Q1 \) or the 25th percentile is determined by solving \( F(Q1) = 0.25 \), which gives \( Q1 = -30 \ln(0.75) \). This represents the kilometer mark below which 25% of the tire lifespans fall.
• \( Q3 \) or the 75th percentile is found by solving \( F(Q3) = 0.75 \), resulting in \( Q3 = -30 \ln(0.25) \). This is the kilometer mark below which 75% of the tire lifespans are found.

The IQR, \( Q3 - Q1 \), thus provides valuable information on the variability of tire lifespan not affected by extreme values or outliers, offering a robust statistical range within the distribution.