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The completion time X for a certain task has cdf F(x) given by

\(\left\{ {\begin{array}{*{20}{c}}{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 0}\\{\frac{{{x^3}}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \le x \le \frac{7}{3}}\\{1 - \frac{1}{2}\left( {\frac{7}{3} - x} \right)\left( {\frac{7}{4} - \frac{3}{4}x} \right)\,\,\,\,\,\,1 \le x \le \frac{7}{3}}\\{1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x > \frac{7}{3}}\end{array}} \right.\)

a. Obtain the pdf f (x) and sketch its graph.

b. Compute\({\bf{P}}\left( {.{\bf{5}} \le {\bf{X}} \le {\bf{2}}} \right)\). c. Compute E(X).

Short Answer

Expert verified

\(\begin{array}{l}(a)\,f(x) = \left\{ {\begin{array}{*{20}{l}}{{x^2}}&{0 \le x < 1}\\{\frac{7}{4} - \left( {\frac{3}{4}} \right)x}&{1 \le x \le \frac{7}{3}}\\0&{{\rm{ otherwise }}}\end{array}} \right.\\(b)\,0.9167\\(c)\,1.213\end{array}\)

Step by step solution

01

Definition of Probability

Probability is a branch of mathematics that studies the probability of a random event occurring. When a coin is tossed into the air, for example, the possible outcomes are Head and Tail.

02

Calculation for the determination of the pdf f(x) in part a.

(a) It is given that X is a random variable denoting the completion time for a certain task. Its CDF F(x) is given as:

Proposition: If X is a continuous rv with pdf f(x) and cdf F(x), then at every x at which the derivative\({F^\prime }(x)\)exists,

\(f(x) = {F^\prime }(x)\)

For intervals\(x < 0\)and\(x > \frac{7}{3}\):

\(f(x) = {F^\prime }(X) = 0\)

For interval\(0 \le x < 1\):

\(f(x) = \frac{d}{{dx}}\left( {\frac{{{x^3}}}{3}} \right) = {x^2}\)

For interval\(1 \le x \le \frac{7}{3}\)

\(\begin{array}{l}f(x) = \frac{d}{{dx}}\left( {1 - \frac{1}{2}\left( {\frac{7}{3} - x} \right)\left( {\frac{7}{4} - \frac{3}{4}x} \right)} \right)\\f(x) = \frac{1}{2}\left( {\frac{7}{4} - \frac{3}{4}x} \right) + \frac{1}{2}\left( {\frac{7}{3} - x} \right)\left( {\frac{3}{4}} \right)\\f(x) = \frac{7}{4} - \frac{3}{4}x\end{array}\)

03

Calculation for the determination of the pdf f(x) in part a.

Hence pdf f(x) can finally be given as:

\(f(x) = \left\{ {\begin{array}{*{20}{l}}{{x^2}}&{0 \le x < 1}\\{\frac{7}{4} - \left( {\frac{3}{4}} \right)x}&{1 \le x \le \frac{7}{3}}\\0&{{\rm{ otherwise }}}\end{array}} \right.\)

04

Calculation for the determination of the probability in part b.

(b) compute\(P(0.5 < X < 2)\)

\(\begin{aligned}P(0.5 < X < 2) &= F(2) - F(0.5) \hfill \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \left[ {1 - \frac{1}{2}\left( {\frac{7}{3} - 2} \right)\left( {\frac{7}{4} - \frac{3}{4} \cdot 2} \right)} \right] - \left[ {\frac{{{{(0.5)}^3}}}{3}} \right] \hfill \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{23}}{{24}} -\frac{1}{{24}} \hfill \\ P(0.5 < X < 2) &= 0.9167 \hfill \\\end{aligned}\)

Proposition: Let\({\rm{X}}\)be a continuous\({\rm{rv}}\)with\({\rm{pdff}}({\rm{x}})\)and\(cdf\;{\rm{F}}({\rm{x}})\). Then for any two numbers a and\({\rm{b}}\)with\(a < b\),

\(P(a \le X \le b) = F(b) - F(a)\)

05

Calculation for the determination of the expected value in part c

(c) The mean value of the given distribution can be given as:

\(\begin{aligned}E(X)&= \int_{ - \infty }^i\cdot f(x) \cdot dx \\ &= \int_0^1 x \cdot {x^2} \cdot dx + \int_1^{7/3} x \cdot \left( {\frac{7}{4} - \frac{3}{4}x} \right) \cdot dx \\&= \int_0^1 {{x^3}} \cdot dx + \int_1^{7/3} {\frac{{7x}}{4}} \cdot dx - \int_1^{7/3} {\frac{{3{x^2}}}{4}} \cdot dx \\&= \left[ {\frac{{{x^4}}}{4}} \right]_0^1 + \left[ {\frac{{7{x^2}}}{8}} \right]_1^{7/3} + \left[ {\frac{{{x^3}}}{4}} \right]_1^{7/3} = {\left[ {\frac{{{1^4}}}{4}} \right]_1} + \left[ {\frac{{343}}{{72}} - \frac{7}{8}} \right] + \left[ {\frac{{343}}{{108}} - \frac{1}{4}} \right] \\&= \frac{1}{4} + \frac{{35}}{9} - \frac{{316}}{{108}} \\E(X) &= 1.213 \\\end{aligned} \)

Definition: The expected or mean value of a continuous\(rv\;X\)with pdf f(x) is

\(\mu = E(X) = \int_{ - \infty }^\infty x \cdot f(x) \cdot dx\)

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Most popular questions from this chapter

The article 鈥淩eliability of Domestic颅Waste Biofilm Reactors鈥 (J. of Envir. Engr., \({\rm{1995: 785--790}}\)) suggests that substrate concentration (mg/cm\({\rm{3}}\)) of influent to a reactor is normally distributed with m \({\rm{5 }}{\rm{.30}}\)and s\({\rm{5 }}{\rm{.06}}\).

a. What is the probability that the concentration exceeds. \({\rm{50}}\)?

b. What is the probability that the concentration is at most. \({\rm{20}}\)?

c. How would you characterize the largest \({\rm{5\% }}\)of all concentration values?

Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate.

\(\begin{array}{l}{\rm{a}}{\rm{. P}}\left( {{\rm{0 拢 Z 拢2}}{\rm{.17}}} \right){\rm{ }}\\{\rm{b}}{\rm{. P}}\left( {{\rm{0拢 Z 拢 1}}} \right){\rm{ }}\\{\rm{c}}{\rm{. P}}\left( {{\rm{ - 2}}{\rm{.50 拢 Z 拢 0}}} \right){\rm{ }}\\{\rm{d}}{\rm{. P}}\left( {{\rm{ - 2}}{\rm{.50 拢 Z 拢 2}}{\rm{.50}}} \right)\\{\rm{ e}}{\rm{. P}}\left( {{\rm{Z 拢 1}}{\rm{.37}}} \right){\rm{ }}\\{\rm{f}}{\rm{. P}}\left( {{\rm{ - 1}}{\rm{.75 拢 Z}}} \right){\rm{ }}\\{\rm{g}}{\rm{. P}}\left( {{\rm{21}}{\rm{.50 拢 Z 拢 2}}{\rm{.00}}} \right){\rm{ }}\\{\rm{h}}{\rm{. P}}\left( {{\rm{1}}{\rm{.37 拢Z 拢 2}}{\rm{.50}}} \right){\rm{ }}\\{\rm{i}}{\rm{. P}}\left( {{\rm{ - 1}}{\rm{.50 拢Z}}} \right){\rm{ }}\\{\rm{j}}{\rm{. P}}\left( {\left| {\rm{Z}} \right|{\rm{ 拢2}}{\rm{.50}}} \right)\end{array}\)

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within \({\rm{2}}\) min after the hour. Let \({\rm{X = }}\)the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of \({\rm{X}}\) is

\({\rm{f(x) = \{ }}\begin{array}{*{20}{c}}{{\rm{k}}{{\rm{x}}^2}}&{{\rm{0}} \le {\rm{x}} \le {\rm{2}}}\\{\rm{0}}&{{\rm{otherwise}}}\end{array}\)

a. Find the value of \({\rm{k}}\) and draw the corresponding density curve. (Hint: Total area under the graph of \({\rm{f(x)}}\) is \({\rm{1}}\).)

b. What is the probability that the lecture ends within \({\rm{1}}\) min of the end of the hour?

c. What is the probability that the lecture continues beyond the hour for between \({\rm{60}}\) and \({\rm{90}}\) sec?

d. What is the probability that the lecture continues for at least \({\rm{90}}\) sec beyond the end of the hour?

Let X 5 the time it takes a read/write head to locate the desired record on a computer disk memory device once the head has been positioned over the correct track. If the disks rotate once every \({\bf{25}}\) milliseconds, a reasonable assumption is that X is uniformly distributed on the interval\(\left( {{\bf{0}},{\rm{ }}{\bf{25}}} \right)\). a. Compute\({\bf{P}}\left( {{\bf{10}} \le {\bf{X}} \le {\bf{20}}} \right)\). b. Compute \({\bf{P}}\left( {{\bf{X}} \le {\bf{10}}} \right)\). c. Obtain the cdf F(X). d. Compute E(X) and \({\sigma _X}\).

The error involved in making a certain measurement is a continuous rv \({\rm{X}}\) with pdf

\({\rm{f(x) = \{ }}\begin{array}{*{20}{c}}{{\rm{.09375(4 - }}{{\rm{x}}^2})}&{{\rm{ - 2}} \le {\rm{x}} \le {\rm{2}}}\\{\rm{0}}&{{\rm{otherwise}}}\end{array}\)

a. Sketch the graph of \({\rm{f(x)}}\).

b. Compute \({\rm{P(X > 0)}}\).

c. Compute \({\rm{P( - 1 < X < 1)}}\).

d. Compute \({\rm{P(X < - }}{\rm{.5 or X > }}{\rm{.5)}}\).

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