Probability and Statistics for Engineering and Science, 9th Edition
Authors: Jay L. Devore
ISBN-13: 978-1305251809
See our solution for Question 2E from Chapter 4 from Devore's Probability and Statistics for Engineering and Science.
Problem 2E
Chapter:
Problem:
Suppose the reaction temperature X (in ?C) in a certain chemical process has a uniform distribution with A = -5 and B = 5. a. Compute P(X <0). b. Compute P(-2.5 < X <2.5). c. ComputeP(-2 ≤ X ≤ 3). d. For k satisfying -5 < k < k + 4 <5, compute P(k < X < k + 4).
Step-by-Step Solution
Step 1
The distribution given is a uniform distribution given by following probability:\[\begin{array}{l}f\left( x \right) = \dfrac{1}{{B - A}}\\ = \dfrac{1}{{5 - \left( { - 5} \right)}}\\ = \dfrac{1}{{10}}\\ = 0.1\end{array}\]
Step 2: (a)
Probability of x less than 0 is given by integration of probability between -5 and 0\[\begin{array}{l}P\left( {X < 0} \right) = \int\limits_{ - 5}^0 {f\left( x \right)dx} \\ = \int\limits_{ - 5}^0 {0.1dx} \\ = 0.1\int\limits_{ - 5}^0 {dx} \\ = 0.1\left[ x \right]_{ - 5}^0\\ = 0.1\left[ {0 + 5} \right]\\ = 0.5\end{array}\]Therefore, \[P\left( {X < 0} \right) = 0.5\]
Step 3: (b)
\[\begin{array}{l}P\left( { - 2.5X < 2.5} \right) = \int\limits_{ - 2.5}^{2.5} {f\left( x \right)dx} \\ = \int\limits_{ - 2.5}^{2.5} {0.1dx} \\ = 0.1\int\limits_{ - 2.5}^{2.5} {dx} \\ = 0.1\left[ x \right]_{ - 2.5}^{2.5}\\ = 0.1\left[ {2.5 + 2.5} \right]\\ = 0.5\end{array}\]Therefore, \[P\left( { - 2.5X < 2.5} \right) = 0.5\]
Step 4: (c)
\[\begin{array}{l}P\left( { - 2X < 3} \right) = \int\limits_{ - 2}^3 {f\left( x \right)dx} \\ = \int\limits_{ - 2}^3 {0.1dx} \\ = 0.1\int\limits_{ - 2}^3 {dx} \\ = 0.1\left[ x \right]_{ - 2}^3\\ = 0.1\left[ {3 + 2} \right]\\ = 0.5\end{array}\]Therefore, \[P\left( { - 2X < 3} \right) = 0.5\]
Step 4: (d)
\[\begin{array}{l}P\left( {k < X < k + 4} \right) = \int\limits_k^{k + 4} {f\left( x \right)dx} \\ = \int\limits_k^{k + 4} {0.1dx} \\ = 0.1\int\limits_k^{k + 4} {dx} \\ = 0.1\left[ x \right]_k^{k + 4}\\ = 0.1\left[ {k + 4 - k} \right]\\ = 0.4\end{array}\]Therefore, \[P\left( {k < X < k + 4} \right) = 0.4\]
The distribution given is a uniform distribution given by following probability:\[\begin{array}{l}f\left( x \right) = \dfrac{1}{{B - A}}\\ = \dfrac{1}{{5 - \left( { - 5} \right)}}\\ = \dfrac{1}{{10}}\\ = 0.1\end{array}\]
Step 2: (a)
Probability of x less than 0 is given by integration of probability between -5 and 0\[\begin{array}{l}P\left( {X < 0} \right) = \int\limits_{ - 5}^0 {f\left( x \right)dx} \\ = \int\limits_{ - 5}^0 {0.1dx} \\ = 0.1\int\limits_{ - 5}^0 {dx} \\ = 0.1\left[ x \right]_{ - 5}^0\\ = 0.1\left[ {0 + 5} \right]\\ = 0.5\end{array}\]Therefore, \[P\left( {X < 0} \right) = 0.5\]
Step 3: (b)
\[\begin{array}{l}P\left( { - 2.5X < 2.5} \right) = \int\limits_{ - 2.5}^{2.5} {f\left( x \right)dx} \\ = \int\limits_{ - 2.5}^{2.5} {0.1dx} \\ = 0.1\int\limits_{ - 2.5}^{2.5} {dx} \\ = 0.1\left[ x \right]_{ - 2.5}^{2.5}\\ = 0.1\left[ {2.5 + 2.5} \right]\\ = 0.5\end{array}\]Therefore, \[P\left( { - 2.5X < 2.5} \right) = 0.5\]
Step 4: (c)
\[\begin{array}{l}P\left( { - 2X < 3} \right) = \int\limits_{ - 2}^3 {f\left( x \right)dx} \\ = \int\limits_{ - 2}^3 {0.1dx} \\ = 0.1\int\limits_{ - 2}^3 {dx} \\ = 0.1\left[ x \right]_{ - 2}^3\\ = 0.1\left[ {3 + 2} \right]\\ = 0.5\end{array}\]Therefore, \[P\left( { - 2X < 3} \right) = 0.5\]
Step 4: (d)
\[\begin{array}{l}P\left( {k < X < k + 4} \right) = \int\limits_k^{k + 4} {f\left( x \right)dx} \\ = \int\limits_k^{k + 4} {0.1dx} \\ = 0.1\int\limits_k^{k + 4} {dx} \\ = 0.1\left[ x \right]_k^{k + 4}\\ = 0.1\left[ {k + 4 - k} \right]\\ = 0.4\end{array}\]Therefore, \[P\left( {k < X < k + 4} \right) = 0.4\]