Problem 2 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X\) be a random variable that takes values in \([0,1]\), and is further given by $$ F(x)=x^{2} \quad \text { for } 0 \leq x \leq 1 . $$ Compute \(\mathrm{P}\left(\frac{1}{2}

Step-by-Step Solution

Answer

\(\mathrm{P}(\frac{1}{2}

Step 1: Identify the Cumulative Distribution Function (CDF)

The problem provides the cumulative distribution function (CDF) for the random variable \( X \). It is given by \( F(x) = x^2 \) for \( 0 \leq x \leq 1 \). This function describes the probability that \( X \) is less than or equal to a certain value \( x \).

Step 2: Calculate F(1/2)

Determine the value of the CDF at \( x = \frac{1}{2} \) to find the probability that \( X \) is less than or equal to \( \frac{1}{2} \). \[ F\left(\frac{1}{2}\right) = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \]

Step 3: Calculate F(3/4)

Next, calculate the value of the CDF at \( x = \frac{3}{4} \) to find the probability that \( X \) is less than or equal to \( \frac{3}{4} \). \[ F\left(\frac{3}{4}\right) = \left( \frac{3}{4} \right)^2 = \frac{9}{16} \]

Step 4: Calculate the Probability Interval

Find \( \mathrm{P}\left(\frac{1}{2}

Key Concepts

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a fundamental concept in probability and statistics. It describes the probability that a random variable takes on a value less than or equal to a specific number. For our example, we have a random variable \( X \) whose CDF is \( F(x) = x^2 \) for \( 0 \leq x \leq 1 \).
The CDF, in simpler terms, is a way to aggregate probabilities on a continuous interval. When you compute \( F(x) \), you are essentially summing up probabilities of all outcomes less than or equal to \( x \).

• In this case, to find the probability that \( X \leq \frac{3}{4} \), we simply calculate \( F(\frac{3}{4}) \).
• Similarly, \( F(\frac{1}{2}) \) gives the probability that \( X \leq \frac{1}{2} \).

The difference between these CDF values gives us the probability of \( X \) falling within a specified interval, which is key to solving problems involving continuous random variables.

Random Variable

A random variable is a variable that takes on numerical values, determined by the outcome of a random phenomenon. In this exercise, \( X \) is a random variable defined over the interval \([0,1]\).
Random variables can be discrete or continuous: the former take on specific values, while the latter can take any value within a range. Our example is a continuous random variable since it can assume any value between 0 and 1.

• The function \( F(x) = x^2 \) represents the CDF of the random variable \( X \), dictating how likely it is that \( X \) will continue a value at or below some threshold \( x \).
• Understanding the nature of a random variable and how it is described by distributions such as CDFs is essential in probability theory and statistics.

The concept of random variables is crucial in translating real-world random phenomena into mathematical frameworks that can be analyzed and understood.

Probability Interval

A probability interval is used to determine the likelihood of a random variable falling within a certain range rather than a specific point. In our exercise, we aim to calculate \( \mathrm{P}(\frac{1}{2}To calculate this, we take the difference between the CDF values at these endpoints, which gives us:
1. Calculate \( F(\frac{3}{4}) \) for the upper limit of the interval.2. Subtract \( F(\frac{1}{2}) \) from it to find the probability confined to this range.

• In numerical terms: \( \mathrm{P}(\frac{1}{2}

This interval doesn't include its lower boundary, \( \frac{1}{2} \), but includes its upper boundary, \( \frac{3}{4} \), which is typical of such range expressions in probability.

Previous

Problem 1

Next

Problem 3

Other exercises in this chapter

Problem 1

Let \(X\) be a continuous random variable with probability density function $$ f(x)= \begin{cases}\frac{3}{4} & \text { for } 0 \leq x \leq 1 \\ \frac{1}{4} & \

View solution

Problem 3

Let a continuous random variable \(X\) be given that takes values in \([0,1]\), and whose distribution function \(F\) satisfies $$ F(x)=2 x^{2}-x^{4} \quad \tex

View solution

Problem 4

Jensen, arriving at a bus stop, just misses the bus. Suppose that he decides to walk if the (next) bus takes longer than 5 minutes to arrive. Suppose also that

View solution

Problem 6

Let \(X\) have an \(\operatorname{Exp}(0.2)\) distribution. Compute \(\mathrm{P}(X>5)\).

View solution

Key Concepts

Cumulative Distribution Function

The Cumulative Distribution Function (CDF) is a fundamental concept in probability and statistics. It describes the probability that a random variable takes on a value less than or equal to a specific number. For our example, we have a random variable \( X \) whose CDF is \( F(x) = x^2 \) for \( 0 \leq x \leq 1 \).
The CDF, in simpler terms, is a way to aggregate probabilities on a continuous interval. When you compute \( F(x) \), you are essentially summing up probabilities of all outcomes less than or equal to \( x \).

• In this case, to find the probability that \( X \leq \frac{3}{4} \), we simply calculate \( F(\frac{3}{4}) \).
• Similarly, \( F(\frac{1}{2}) \) gives the probability that \( X \leq \frac{1}{2} \).

The difference between these CDF values gives us the probability of \( X \) falling within a specified interval, which is key to solving problems involving continuous random variables.

Random Variable

A random variable is a variable that takes on numerical values, determined by the outcome of a random phenomenon. In this exercise, \( X \) is a random variable defined over the interval \([0,1]\).
Random variables can be discrete or continuous: the former take on specific values, while the latter can take any value within a range. Our example is a continuous random variable since it can assume any value between 0 and 1.

• The function \( F(x) = x^2 \) represents the CDF of the random variable \( X \), dictating how likely it is that \( X \) will continue a value at or below some threshold \( x \).
• Understanding the nature of a random variable and how it is described by distributions such as CDFs is essential in probability theory and statistics.

The concept of random variables is crucial in translating real-world random phenomena into mathematical frameworks that can be analyzed and understood.

Probability Interval

A probability interval is used to determine the likelihood of a random variable falling within a certain range rather than a specific point. In our exercise, we aim to calculate \( \mathrm{P}(\frac{1}{2}To calculate this, we take the difference between the CDF values at these endpoints, which gives us:
1. Calculate \( F(\frac{3}{4}) \) for the upper limit of the interval.2. Subtract \( F(\frac{1}{2}) \) from it to find the probability confined to this range.

• In numerical terms: \( \mathrm{P}(\frac{1}{2}

This interval doesn't include its lower boundary, \( \frac{1}{2} \), but includes its upper boundary, \( \frac{3}{4} \), which is typical of such range expressions in probability.