et X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following F(x) = Use the cdf to obtain the following. (If necessary, round your answer to four decimal places.) (a) Calculate P(X ≤ 3). 0.36 (c) Calculate P(X> 3.5). 51 0 (b) Calculate P(2.5 ≤ x ≤ 3). 11 { 25 1 2 (d) What is the median checkout duration ? [solve 0.5 = F(μ)]. 3.535 X x x < 0 0
et X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following F(x) = Use the cdf to obtain the following. (If necessary, round your answer to four decimal places.) (a) Calculate P(X ≤ 3). 0.36 (c) Calculate P(X> 3.5). 51 0 (b) Calculate P(2.5 ≤ x ≤ 3). 11 { 25 1 2 (d) What is the median checkout duration ? [solve 0.5 = F(μ)]. 3.535 X x x < 0 0
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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I can't figure out parts D and F. For part D I did 25 times .5 = mu tilde2 and I got 3.536 and it said it was wrong. I also tried 3.535 and that was also wrong, so I don't know what I'm doing wrong. For part F I did some steps I can't explain in here. Can you please help me with parts D and F?
![**Cumulative Distribution Function (CDF) and Related Calculations**
Let \( X \) denote the amount of time a book on two-hour reserve is actually checked out. The cumulative distribution function (CDF) is given by:
\[
F(x) =
\begin{cases}
0 & x < 0 \\
\frac{x^2}{25} & 0 \le x < 5 \\
1 & 5 \le x
\end{cases}
\]
**Use the CDF to obtain the following: (If necessary, round your answer to four decimal places.)**
**(a) Calculate \( P(X \leq 3) \).**
- Answer: 0.36 ✓
**(b) Calculate \( P(2.5 \leq X \leq 3) \).**
- Answer: 0.11 ✓
**(c) Calculate \( P(X > 3.5) \).**
- Answer: 0.51 ✓
**(d) What is the median checkout duration \(\tilde{\mu}\)? [Solve \( 0.5 = F(\tilde{\mu}) \)].**
- Proposed Answer: 3.535 ✗
**(e) Obtain the density function \( f(x) \).**
The density function \( f(x) = F'(x) \) is provided by:
\[
f(x) =
\begin{cases}
\frac{2}{25}x & 0 \le x < 5 \\
0 & \text{otherwise}
\end{cases}
\]
**(f) Calculate \( E(X) \).**
- Proposed Answer: 3.333 ✗
(Graphs or diagrams are not present in the image and thus are not explained here.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff76fb667-0622-451e-86fc-d3866765f8e5%2F78f98823-05f7-4213-8174-1812c9c393ae%2Fe38oc8f_processed.png&w=3840&q=75)
Transcribed Image Text:**Cumulative Distribution Function (CDF) and Related Calculations**
Let \( X \) denote the amount of time a book on two-hour reserve is actually checked out. The cumulative distribution function (CDF) is given by:
\[
F(x) =
\begin{cases}
0 & x < 0 \\
\frac{x^2}{25} & 0 \le x < 5 \\
1 & 5 \le x
\end{cases}
\]
**Use the CDF to obtain the following: (If necessary, round your answer to four decimal places.)**
**(a) Calculate \( P(X \leq 3) \).**
- Answer: 0.36 ✓
**(b) Calculate \( P(2.5 \leq X \leq 3) \).**
- Answer: 0.11 ✓
**(c) Calculate \( P(X > 3.5) \).**
- Answer: 0.51 ✓
**(d) What is the median checkout duration \(\tilde{\mu}\)? [Solve \( 0.5 = F(\tilde{\mu}) \)].**
- Proposed Answer: 3.535 ✗
**(e) Obtain the density function \( f(x) \).**
The density function \( f(x) = F'(x) \) is provided by:
\[
f(x) =
\begin{cases}
\frac{2}{25}x & 0 \le x < 5 \\
0 & \text{otherwise}
\end{cases}
\]
**(f) Calculate \( E(X) \).**
- Proposed Answer: 3.333 ✗
(Graphs or diagrams are not present in the image and thus are not explained here.)
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I don't know how to do parts G and H. Can you please help me?
![Let \( X \) denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cumulative distribution function (cdf) is the following:
\[
F(x) =
\begin{cases}
0 & x < 0 \\
\frac{x^2}{25} & 0 \leq x < 5 \\
1 & 5 \leq x
\end{cases}
\]
Use the cdf to obtain the following. (If necessary, round your answer to four decimal places.)
**(a)** Calculate \( P(X \leq 3) \).
\[
0.36 \quad \checkmark
\]
**(b)** Calculate \( P(2.5 \leq X \leq 3) \).
\[
0.11 \quad \checkmark
\]
**(c)** Calculate \( P(X > 3.5) \).
\[
0.51 \quad \checkmark
\]
**(d)** What is the median checkout duration \( \tilde{\mu} \)? [Solve \( 0.5 = F(\tilde{\mu}) \)].
\[
3.5355 \quad \checkmark
\]
**(e)** Obtain the density function \( f(x) \).
\[
f(x) = F'(x)
\]
\[
f(x) =
\begin{cases}
\frac{2x}{25} & 0 \leq x < 5 \\
0 & \text{otherwise}
\end{cases}
\quad \checkmark
\]
**(f)** Calculate \( E(X) \).
\[
3.3333 \quad \checkmark
\]
**(g)** Calculate \( V(X) \) and \( \sigma_X \).](https://content.bartleby.com/qna-images/question/f76fb667-0622-451e-86fc-d3866765f8e5/d4e11c0b-d957-4f99-8823-5c42d0e7860f/z2l86g_processed.png)
Transcribed Image Text:Let \( X \) denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cumulative distribution function (cdf) is the following:
\[
F(x) =
\begin{cases}
0 & x < 0 \\
\frac{x^2}{25} & 0 \leq x < 5 \\
1 & 5 \leq x
\end{cases}
\]
Use the cdf to obtain the following. (If necessary, round your answer to four decimal places.)
**(a)** Calculate \( P(X \leq 3) \).
\[
0.36 \quad \checkmark
\]
**(b)** Calculate \( P(2.5 \leq X \leq 3) \).
\[
0.11 \quad \checkmark
\]
**(c)** Calculate \( P(X > 3.5) \).
\[
0.51 \quad \checkmark
\]
**(d)** What is the median checkout duration \( \tilde{\mu} \)? [Solve \( 0.5 = F(\tilde{\mu}) \)].
\[
3.5355 \quad \checkmark
\]
**(e)** Obtain the density function \( f(x) \).
\[
f(x) = F'(x)
\]
\[
f(x) =
\begin{cases}
\frac{2x}{25} & 0 \leq x < 5 \\
0 & \text{otherwise}
\end{cases}
\quad \checkmark
\]
**(f)** Calculate \( E(X) \).
\[
3.3333 \quad \checkmark
\]
**(g)** Calculate \( V(X) \) and \( \sigma_X \).
![### Educational Content on Probability and Statistics
#### Problem Summary and Solutions
**(c) Calculate \( P(X > 3.5) \):**
The probability that the random variable \( X \) is greater than 3.5 is 0.51.
**(d) Median Checkout Duration \( \tilde{\mu} \):**
To find the median checkout duration, solve \( 0.5 = F(\tilde{\mu}) \). The median duration is 3.5355.
**(e) Density Function \( f(x) \):**
The density function can be obtained by taking the derivative of the cumulative distribution function \( F(x) \):
\[
f(x) = F'(x) =
\begin{cases}
\frac{2}{25} x & \text{for } 0 \leq x < 5 \\
0 & \text{otherwise}
\end{cases}
\]
**(f) Expected Value \( E(X) \):**
The expected value of the random variable \( X \) is 3.3333.
**(g) Variance \( V(X) \) and Standard Deviation \( \sigma_X \):**
The variance \( V(X) \) and standard deviation \( \sigma_X \) are not provided in the image and may need to be calculated using additional information.
**(h) Expected Charge \( E[h(X)] \):**
If the borrower is charged an amount \( h(X) = X^2 \) based on the checkout duration \( X \), the expected charge \( E[h(X)] \) is not provided in the image and requires additional calculation based on the given density function.
### Explanation of Concepts
- **Cumulative Distribution Function (CDF):** The function \( F(x) \) represents the probability that a random variable \( X \) will take a value less than or equal to \( x \).
- **Probability Density Function (PDF):** The derivative of the CDF, \( f(x) \), provides the likelihood of the random variable \( X \) being exactly equal to \( x \).
- **Expected Value:** The mean or average value of the random variable, calculated by integrating the product of the variable and its PDF over its entire range.
- **Variance and Standard Deviation:** Measures of the spread of the random variable's values. Variance is the expectation of](https://content.bartleby.com/qna-images/question/f76fb667-0622-451e-86fc-d3866765f8e5/d4e11c0b-d957-4f99-8823-5c42d0e7860f/n5israi_processed.png)
Transcribed Image Text:### Educational Content on Probability and Statistics
#### Problem Summary and Solutions
**(c) Calculate \( P(X > 3.5) \):**
The probability that the random variable \( X \) is greater than 3.5 is 0.51.
**(d) Median Checkout Duration \( \tilde{\mu} \):**
To find the median checkout duration, solve \( 0.5 = F(\tilde{\mu}) \). The median duration is 3.5355.
**(e) Density Function \( f(x) \):**
The density function can be obtained by taking the derivative of the cumulative distribution function \( F(x) \):
\[
f(x) = F'(x) =
\begin{cases}
\frac{2}{25} x & \text{for } 0 \leq x < 5 \\
0 & \text{otherwise}
\end{cases}
\]
**(f) Expected Value \( E(X) \):**
The expected value of the random variable \( X \) is 3.3333.
**(g) Variance \( V(X) \) and Standard Deviation \( \sigma_X \):**
The variance \( V(X) \) and standard deviation \( \sigma_X \) are not provided in the image and may need to be calculated using additional information.
**(h) Expected Charge \( E[h(X)] \):**
If the borrower is charged an amount \( h(X) = X^2 \) based on the checkout duration \( X \), the expected charge \( E[h(X)] \) is not provided in the image and requires additional calculation based on the given density function.
### Explanation of Concepts
- **Cumulative Distribution Function (CDF):** The function \( F(x) \) represents the probability that a random variable \( X \) will take a value less than or equal to \( x \).
- **Probability Density Function (PDF):** The derivative of the CDF, \( f(x) \), provides the likelihood of the random variable \( X \) being exactly equal to \( x \).
- **Expected Value:** The mean or average value of the random variable, calculated by integrating the product of the variable and its PDF over its entire range.
- **Variance and Standard Deviation:** Measures of the spread of the random variable's values. Variance is the expectation of
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