Let X denote the amount of time a book on two-hour reserve is actually checked out, and suppose the cdf is the following. 0 x < 0 P-f x² F(x) 0≤x < 4 16 1 4 SX Use the cdf to obtain the following. (If necessary, round your answer to four decimal places.) (a) Calculate P(X ≤ 3). 0.5625 (b) Calculate P(2.5 ≤ x ≤ 3). 0.1719 (c) Calculate P(X> 3.5). 0.2344 (d) What is the median checkout duration ? [solve 0.5 = F(μ)]. (e) Obtain the density function f(x). f(x) = F'(x) 0≤x < 4 otherwise (f) Calculate E(X). (g) Calculate V(X) and a V(X) %% (h) If the borrower is charged an amount h(x) = x² when checkout duration is X, compute the expected charge E[h(X)].

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P17. Please only answer the questions to the left of the red X. Thank you

### Probability and Statistics: Continuous Random Variables

#### Distribution Functions and Calculations

Let \( X \) denote the amount of time a book on two-hour reserve is actually checked out. Suppose the cumulative distribution function (CDF) is given by:

\[ F(x) = 
\begin{cases} 
0 & x < 0 \\
\frac{x^2}{16} & 0 \leq x < 4 \\
1 & 4 \leq x 
\end{cases} 
\]

Using this CDF, we can derive various properties and statistics of the random variable \( X \).

#### (a) Calculate \( P(X \leq 3) \).

To find \( P(X \leq 3) \), we evaluate the CDF at \( x = 3 \):
\[ F(3) = \frac{3^2}{16} = \frac{9}{16} \approx 0.5625 \]

#### (b) Calculate \( P(2.5 \leq X \leq 3) \).

This probability is given by the difference \( F(3) - F(2.5) \):
\[ F(2.5) = \frac{2.5^2}{16} = \frac{6.25}{16} \approx 0.3906 \]
\[ P(2.5 \leq X \leq 3) = F(3) - F(2.5) = 0.5625 - 0.3906 \approx 0.1719 \]

#### (c) Calculate \( P(X > 3.5) \).

This probability is given by \( 1 - F(3.5) \):
\[ F(3.5) = \frac{3.5^2}{16} = \frac{12.25}{16} \approx 0.7656 \]
\[ P(X > 3.5) = 1 - 0.7656 \approx 0.2344 \]

#### (d) What is the median checkout duration \( \tilde{x} \)? 

The median checkout duration is the value \( \tilde{x} \) such that \( F(\tilde{x}) = 0.5 \):

\[ F(\tilde{x}) = \frac{\tilde{x}^2}{16} =
Transcribed Image Text:### Probability and Statistics: Continuous Random Variables #### Distribution Functions and Calculations Let \( X \) denote the amount of time a book on two-hour reserve is actually checked out. Suppose the cumulative distribution function (CDF) is given by: \[ F(x) = \begin{cases} 0 & x < 0 \\ \frac{x^2}{16} & 0 \leq x < 4 \\ 1 & 4 \leq x \end{cases} \] Using this CDF, we can derive various properties and statistics of the random variable \( X \). #### (a) Calculate \( P(X \leq 3) \). To find \( P(X \leq 3) \), we evaluate the CDF at \( x = 3 \): \[ F(3) = \frac{3^2}{16} = \frac{9}{16} \approx 0.5625 \] #### (b) Calculate \( P(2.5 \leq X \leq 3) \). This probability is given by the difference \( F(3) - F(2.5) \): \[ F(2.5) = \frac{2.5^2}{16} = \frac{6.25}{16} \approx 0.3906 \] \[ P(2.5 \leq X \leq 3) = F(3) - F(2.5) = 0.5625 - 0.3906 \approx 0.1719 \] #### (c) Calculate \( P(X > 3.5) \). This probability is given by \( 1 - F(3.5) \): \[ F(3.5) = \frac{3.5^2}{16} = \frac{12.25}{16} \approx 0.7656 \] \[ P(X > 3.5) = 1 - 0.7656 \approx 0.2344 \] #### (d) What is the median checkout duration \( \tilde{x} \)? The median checkout duration is the value \( \tilde{x} \) such that \( F(\tilde{x}) = 0.5 \): \[ F(\tilde{x}) = \frac{\tilde{x}^2}{16} =
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