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The cumulative distribution function (CDF) of a random variable provides a way to understand the probability that the variable will take a value less than or equal to a specific value. For a given probability density function (PDF), the CDF is derived by integrating the PDF over its domain. For the PDF given in the exercise, the CDF is expressed as:\[ F(x) = \int_0^x 90t^8(1-t) \, dt \]This formula represents the accumulation of probabilities from 0 to whatever specific point x you choose within the range of 0 to 1. Graphically, the CDF is an increasing function that starts at 0 (when x is 0) and asymptotically approaches 1 as x reaches its upper bound. When plotting the CDF, you should see a smooth curve that confirms how probabilities accumulate along the interval. Calculating specific probabilities using the CDF involves evaluating it at a specific point, as shown in problems parts b and c of the exercise.

The expected value, or mean, of a random variable is a measure of its central tendency. It's akin to the long-term average value that you would expect over many observations of the random variable. For a continuous random variable defined over a probability density function like the one in this exercise, the expected value is determined by the following integral:\[ E(X) = \int_0^1 x \cdot f(x) \, dx \]Substituting the function given, you'll evaluate:\[ E(X) = \int_0^1 x \cdot 90x^8(1-x) \, dx \]This calculation provides E(X), which tells us the 'center' or the balancing point of the distribution. The expected value is essential because it helps to summarize the entire probability distribution into a single, meaningful quantity. It allows you to make various predictions and decisions based on this central value.

Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of values. It is pivotal in understanding the spread of a distribution around its mean (or expected value). For a random variable with a known PDF, the process to find the standard deviation involves finding the variance first and then taking the square root.

• First, find the expected value of the square, \( E(X^2) = \int_0^1 x^2 \cdot f(x) \, dx \).
• Compute the variance as \( \sigma_X^2 = E(X^2) - [E(X)]^2 \).
• Then, the standard deviation is \( \sigma_X = \sqrt{\sigma_X^2} \).

Standard deviation provides an understanding of how much individual values can deviate from the mean. A higher standard deviation means data is more spread out, while a lower standard deviation indicates data is closer to the mean.

Percentiles are values below which a certain percent of observations fall. They are an efficient way to describe the distribution of data. The exercise discusses finding the 75th percentile, which is a common task to assess where most values lie under the cumulative distribution plot.
To locate the 75th percentile, you solve the equation:\[ F(x) = 0.75 \]This involves the inverse of the CDF to find the particular x that makes the cumulative probability reach 75%. In simple terms, the 75th percentile marks the value below which 75% of the observations fall. Understanding percentiles is helpful for comparative analysis, seeing data distribution, and making informed decisions based on ranking criteria.