Suppose that the continuous random variable X has a cumulative distribution function given by 0, if x < √2 x² -2 if √2 < x < √3 if √3 < x. 1, F(x)= = (a) Find the smallest interval [a, b] such that of P(a ≤ x ≤ b) = 1. (b) Find P(X= 1.6). (c) Find P(1 < X < 1). (d) Find the probability density function of X.
Suppose that the continuous random variable X has a cumulative distribution function given by 0, if x < √2 x² -2 if √2 < x < √3 if √3 < x. 1, F(x)= = (a) Find the smallest interval [a, b] such that of P(a ≤ x ≤ b) = 1. (b) Find P(X= 1.6). (c) Find P(1 < X < 1). (d) Find the probability density function of X.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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![**Continuous Random Variable and Its Properties**
Let's consider a continuous random variable \( X \) with its cumulative distribution function (CDF) \( F(x) \) defined as follows:
\[ F(x) =
\begin{cases}
0 & \text{if } x < \sqrt{2} \\
x^2 - 2 & \text{if } \sqrt{2} \leq x < \sqrt{3} \\
1 & \text{if } \sqrt{3} \leq x
\end{cases}
\]
We need to solve the following problems:
(a) **Find the smallest interval \([a, b]\) such that \(P(a \leq X \leq b) = 1\).**
**Solution:**
To satisfy \( P(a \leq X \leq b) = 1 \), the interval \([a, b]\) must include all the probability mass of the random variable \( X \). According to the CDF provided, \( F(x) \) is 0 for \( x < \sqrt{2} \) and becomes 1 at \( x = \sqrt{3} \). Thus, the smallest interval \([a, b]\) such that \( P(a \leq X \leq b) = 1 \) is:
\[ \left[ \sqrt{2}, \sqrt{3} \right] \]
(b) **Find \( P(X = 1.6) \).**
**Solution:**
For continuous random variables, the probability of the variable taking any specific exact value is always 0. Therefore,
\[ P(X = 1.6) = 0 \]
(c) **Find \( P(1 \leq X \leq \frac{3}{2}) \).**
**Solution:**
To find \( P(1 \leq X \leq \frac{3}{2}) \), we need to evaluate the CDF at the endpoints and subtract:
- At \( X = 1 \): Since \( 1 < \sqrt{2} \), \( F(1) = 0 \)
- At \( X = \frac{3}{2} \): Since \( \sqrt{2} \leq \frac{3}{2} < \sqrt{3} \), \( F\left](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e7c908-505f-4c35-8ac4-e541f2ab69b2%2F82f8d71c-2a70-4b94-b5c0-7ef4814b7bc9%2Fc58ml4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Continuous Random Variable and Its Properties**
Let's consider a continuous random variable \( X \) with its cumulative distribution function (CDF) \( F(x) \) defined as follows:
\[ F(x) =
\begin{cases}
0 & \text{if } x < \sqrt{2} \\
x^2 - 2 & \text{if } \sqrt{2} \leq x < \sqrt{3} \\
1 & \text{if } \sqrt{3} \leq x
\end{cases}
\]
We need to solve the following problems:
(a) **Find the smallest interval \([a, b]\) such that \(P(a \leq X \leq b) = 1\).**
**Solution:**
To satisfy \( P(a \leq X \leq b) = 1 \), the interval \([a, b]\) must include all the probability mass of the random variable \( X \). According to the CDF provided, \( F(x) \) is 0 for \( x < \sqrt{2} \) and becomes 1 at \( x = \sqrt{3} \). Thus, the smallest interval \([a, b]\) such that \( P(a \leq X \leq b) = 1 \) is:
\[ \left[ \sqrt{2}, \sqrt{3} \right] \]
(b) **Find \( P(X = 1.6) \).**
**Solution:**
For continuous random variables, the probability of the variable taking any specific exact value is always 0. Therefore,
\[ P(X = 1.6) = 0 \]
(c) **Find \( P(1 \leq X \leq \frac{3}{2}) \).**
**Solution:**
To find \( P(1 \leq X \leq \frac{3}{2}) \), we need to evaluate the CDF at the endpoints and subtract:
- At \( X = 1 \): Since \( 1 < \sqrt{2} \), \( F(1) = 0 \)
- At \( X = \frac{3}{2} \): Since \( \sqrt{2} \leq \frac{3}{2} < \sqrt{3} \), \( F\left
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