Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 3E from Chapter 4 from Devore's Probability and Statistics for Engineering and Science.
Problem 3E
Chapter:
Problem:
The error involved in making a certain measurement is a continuous rv X with pdf ... a. Sketch the graph of f ( x ). b. Compute P(X > 0). c. Compute P(-1 < X < 1). d. ComputeP(X <-.5 or X > .5)
Step-by-Step Solution
Step 1
The probability distribution function of X (error involved in making a certain measurement)\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{0.09375\left( {4 - {x^2}} \right)}&{ - 2 \le x \le 2}\\0&{{\rm{Otherwise}}}\end{array}} \right.\]
Step 2: (a)
The plot of the function along with some values is given below:https://imgur.com/9EH4Lr5
Step 3: (b)
\[\begin{array}{l}P\left( {X > 0} \right) = \int\limits_0^2 {f\left( x \right)dx} \\ = \int\limits_0^2 {0.09375\left( {4 - {x^2}} \right)dx} \\ = 0.09375\int\limits_0^2 {\left( {4 - {x^2}} \right)dx} \\ = 0.09375\left[ {4x - \frac{{{x^3}}}{3}} \right]_0^2\\ = 0.09375\left[ {4\left( 2 \right) - \frac{{{2^3}}}{3}} \right]\\ = 0.09375\left[ {8 - \frac{8}{3}} \right]\\ = 0.5\end{array}\]Therefore, \[P\left( {X > 0} \right) = 0.5\]
Step 4: (c)
\[\begin{array}{l}P\left( { - 1 < X < 1} \right) = \int\limits_{ - 1}^1 {f\left( x \right)dx} \\ = \int\limits_{ - 1}^1 {0.09375\left( {4 - {x^2}} \right)dx} \\ = 0.09375\int\limits_{ - 1}^1 {\left( {4 - {x^2}} \right)dx} \\ = 0.09375\left[ {4x - \frac{{{x^3}}}{3}} \right]_{ - 1}^1\\ = 0.09375\left[ {4\left( 1 \right) - \frac{{{1^3}}}{3} - 4\left( { - 1} \right) - \frac{{{{\left( { - 1} \right)}^3}}}{3}} \right]\\ = 0.09375\left[ {8 - \frac{2}{3}} \right]\\ = 0.688\end{array}\]Therefore, \[P\left( { - 1 < X < 1} \right) = 0.688\]
Step 4: (d)
\[\begin{array}{l}P\left( {X < - 0.5\,{\rm{or}}\,\,X > 0.5} \right) = \int\limits_{ - 2}^{ - 0.5} {f\left( x \right)dx} + \int\limits_{0.5}^2 {f\left( x \right)dx} \\ = \int\limits_{ - 2}^{ - 0.5} {0.09375\left( {4 - {x^2}} \right)dx} + \int\limits_{0.5}^2 {0.09375\left( {4 - {x^2}} \right)dx} \\ = 0.09375\int\limits_{ - 2}^{ - 0.5} {\left( {4 - {x^2}} \right)dx} + 0.09375\int\limits_{0.5}^2 {\left( {4 - {x^2}} \right)dx} \\ = 0.09375\left[ {4x - \frac{{{x^3}}}{3}} \right]_{ - 2}^{ - 0.5} + 0.09375\left[ {4x - \frac{{{x^3}}}{3}} \right]_{0.5}^2\\ = 0.09375\left[ {4\left( { - 0.5} \right) - \frac{{{{\left( { - 0.5} \right)}^3}}}{3} - 4\left( { - 2} \right) - \frac{{{{\left( { - 2} \right)}^3}}}{3}} \right] + 0.09375\left[ {4\left( 2 \right) - \frac{{{2^3}}}{3} - 4\left( {0.5} \right) - \frac{{{{\left( {0.5} \right)}^3}}}{3}} \right]\\ = 0.3164 + 0.3164\\ = 0.6328\end{array}\]Therefore, \[P\left( {X < - 0.5\,{\rm{or}}\,\,X > 0.5} \right) = 0.6328\]
The probability distribution function of X (error involved in making a certain measurement)\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{0.09375\left( {4 - {x^2}} \right)}&{ - 2 \le x \le 2}\\0&{{\rm{Otherwise}}}\end{array}} \right.\]
Step 2: (a)
The plot of the function along with some values is given below:https://imgur.com/9EH4Lr5
Step 3: (b)
\[\begin{array}{l}P\left( {X > 0} \right) = \int\limits_0^2 {f\left( x \right)dx} \\ = \int\limits_0^2 {0.09375\left( {4 - {x^2}} \right)dx} \\ = 0.09375\int\limits_0^2 {\left( {4 - {x^2}} \right)dx} \\ = 0.09375\left[ {4x - \frac{{{x^3}}}{3}} \right]_0^2\\ = 0.09375\left[ {4\left( 2 \right) - \frac{{{2^3}}}{3}} \right]\\ = 0.09375\left[ {8 - \frac{8}{3}} \right]\\ = 0.5\end{array}\]Therefore, \[P\left( {X > 0} \right) = 0.5\]
Step 4: (c)
\[\begin{array}{l}P\left( { - 1 < X < 1} \right) = \int\limits_{ - 1}^1 {f\left( x \right)dx} \\ = \int\limits_{ - 1}^1 {0.09375\left( {4 - {x^2}} \right)dx} \\ = 0.09375\int\limits_{ - 1}^1 {\left( {4 - {x^2}} \right)dx} \\ = 0.09375\left[ {4x - \frac{{{x^3}}}{3}} \right]_{ - 1}^1\\ = 0.09375\left[ {4\left( 1 \right) - \frac{{{1^3}}}{3} - 4\left( { - 1} \right) - \frac{{{{\left( { - 1} \right)}^3}}}{3}} \right]\\ = 0.09375\left[ {8 - \frac{2}{3}} \right]\\ = 0.688\end{array}\]Therefore, \[P\left( { - 1 < X < 1} \right) = 0.688\]
Step 4: (d)
\[\begin{array}{l}P\left( {X < - 0.5\,{\rm{or}}\,\,X > 0.5} \right) = \int\limits_{ - 2}^{ - 0.5} {f\left( x \right)dx} + \int\limits_{0.5}^2 {f\left( x \right)dx} \\ = \int\limits_{ - 2}^{ - 0.5} {0.09375\left( {4 - {x^2}} \right)dx} + \int\limits_{0.5}^2 {0.09375\left( {4 - {x^2}} \right)dx} \\ = 0.09375\int\limits_{ - 2}^{ - 0.5} {\left( {4 - {x^2}} \right)dx} + 0.09375\int\limits_{0.5}^2 {\left( {4 - {x^2}} \right)dx} \\ = 0.09375\left[ {4x - \frac{{{x^3}}}{3}} \right]_{ - 2}^{ - 0.5} + 0.09375\left[ {4x - \frac{{{x^3}}}{3}} \right]_{0.5}^2\\ = 0.09375\left[ {4\left( { - 0.5} \right) - \frac{{{{\left( { - 0.5} \right)}^3}}}{3} - 4\left( { - 2} \right) - \frac{{{{\left( { - 2} \right)}^3}}}{3}} \right] + 0.09375\left[ {4\left( 2 \right) - \frac{{{2^3}}}{3} - 4\left( {0.5} \right) - \frac{{{{\left( {0.5} \right)}^3}}}{3}} \right]\\ = 0.3164 + 0.3164\\ = 0.6328\end{array}\]Therefore, \[P\left( {X < - 0.5\,{\rm{or}}\,\,X > 0.5} \right) = 0.6328\]