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Magazine Chapter 4: Q8E (page 147)
In commuting to work, a professor must first get on a bus near her house and then transfer to a second bus. If the waiting time (in minutes) at each stop has a uniform distribution with \({\rm{A = 0}}\) and \({\rm{B = 5}}\), then it can be shown that the total waiting time \({\rm{Y}}\) has the pdf
\({\rm{f(x) = \{ }}\begin{array}{*{20}{c}}{\frac{{\rm{1}}}{{{\rm{25}}}}{\rm{y}}}\\{\frac{{\rm{2}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{25}}}}{\rm{y}}}\\{\rm{0}}\end{array}\begin{array}{*{20}{c}}{{\rm{0}} \le {\rm{y < 5}}}\\{{\rm{5}} \le {\rm{y}} \le {\rm{10}}}\\{{\rm{y < 0ory > 10}}}\end{array}\)
a. Sketch a graph of the pdf of \({\rm{Y}}\).
b. Verify that \(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(y)dy = 1}}} \).
c. What is the probability that total waiting time is at most \(3\) min?
d. What is the probability that total waiting time is at most \(8\) min?
e. What is the probability that total waiting time is between \(3\) and \(8\) min?
f. What is the probability that total waiting time is either less than \(2\) min or more than \(6\) min?
Short Answer
(a) The graph for the pdf of\({\rm{Y}}\)is -
(b) It is verified that \(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(y)dy = 1}}} \).
(c) The probability that total waiting time is at most\({\rm{3}}\)min is\({\rm{0}}{\rm{.18}}\).
(d) The probability that total waiting time is at most\({\rm{8}}\)min is\({\rm{0}}{\rm{.92}}\).
(e) The probability that total waiting time is between\({\rm{3}}\)and\({\rm{8}}\)min is\({\rm{0}}{\rm{.74}}\).
(f) The probability that total waiting time is either less than \({\rm{2}}\) min or more than \({\rm{6}}\) min is \({\rm{0}}{\rm{.4}}\).
Step by step solution
Concept Introduction
Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.
The graph for pdf
(a)
The graph of pdfis given below 鈥�
Therefore, the graph is obtained.
Verification of \(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(y)dy = 1}}} \)
(b)
The pdf of\({\rm{f(y)}}\)is given as 鈥�
\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{\rm{y}}}{{{\rm{25}}}}}&{{\rm{0}} \le {\rm{Y}} \le {\rm{5}}}\\{\frac{{\rm{2}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{y}}}{{{\rm{25}}}}}&{{\rm{5 < Y}} \le {\rm{10}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)
From this pdf, it can be written 鈥�
\(\begin{aligned}\int_\infty ^\infty {\rm{f}} {\rm{(y)}} \cdot dy &= \int_{\rm{0}}^{\rm{5}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy + }}\int_{\rm{5}}^{{\rm{10}}} {\left( {\frac{{\rm{2}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{y}}}{{{\rm{25}}}}} \right)} \cdot {\rm{dy}}\\ &= \int_{\rm{0}}^{\rm{5}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy + }}\int_{\rm{5}}^{{\rm{10}}} {\frac{{\rm{2}}}{{\rm{5}}}} \cdot {\rm{dy - }}\int_{\rm{5}}^{{\rm{10}}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\int_{\rm{0}}^{\rm{5}} {\rm{y}} \cdot {\rm{dy + }}\frac{{\rm{2}}}{{\rm{5}}}\int_{\rm{5}}^{{\rm{10}}} {\rm{d}} {\rm{y - }}\frac{{\rm{1}}}{{{\rm{25}}}}\int_{\rm{5}}^{{\rm{10}}} {\rm{y}} \cdot {\rm{dy}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{0}}^{\rm{5}}{\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(y)}}_{\rm{5}}^{{\rm{10}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{5}}^{{\rm{10}}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{5}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{0}}^{\rm{2}}}}}{{\rm{2}}}} \right){\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(10 - 5) - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{5}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\&= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{\rm{25}}}}{{\rm{2}}}} \right){\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(5) - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{\rm{75}}}}{{\rm{2}}}} \right)\\&= \frac{{\rm{1}}}{{\rm{2}}}{\rm{ + 2 - }}\frac{{\rm{3}}}{{\rm{2}}}\\\int_\infty ^\infty {\rm{f}} {\rm{(y)}} \cdot dy &= 1\end{aligned}\)
This could also be proved by using the following property of any pdf 鈥�
\(\int_\infty ^\infty {\rm{f}} {\rm{(y)}} \cdot {\rm{dy = Area under graph of f(y)}}\)
Since the graph of \({\rm{f(y)}}\) is a triangle with base\({\rm{ = 10}}\) and height\({\rm{ = 0}}{\rm{.2}}\), hence 鈥�
\(\int_\infty ^\infty {\rm{f}} {\rm{(y)}} \cdot {\rm{dy = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ \times 0}}{\rm{.2 \times 10 = 1}}\)
Therefore, the verification is done.
Finding the Probability
(c)
The probability that the total waiting time is at most \({\rm{3}}\) min is given as \({\rm{P(Y}} \le {\rm{3)}}\). Using the pdf \({\rm{f(y)}}\), can be written as 鈥�
\(\begin{aligned}{\rm{P(Y}} \le 3) &= \int_{\rm{0}}^{\rm{3}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\int_{\rm{0}}^{\rm{3}} {\rm{y}} \cdot {\rm{dy}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{0}}^{\rm{3}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{3}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{0}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{\rm{9}}}{{\rm{2}}}} \right)\\&= \frac{{\rm{9}}}{{{\rm{50}}}}\\{\rm{P(Y}} \le 3) &= 0 {\rm{.18}}\end{aligned}\)
Therefore, the value is obtained as \({\rm{0}}{\rm{.18}}\).
Finding the Probability
(d)
The probability that the total waiting time is at most \({\rm{8}}\) min is given as \({\rm{P(Y}} \le {\rm{8)}}\). Using the pdf \({\rm{f(y)}}\), can be written as 鈥�
\(\begin{aligned}{\rm{P(Y}} \le 8) &= \int_{\rm{0}}^{\rm{5}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy + }}\int_{\rm{5}}^{\rm{8}} {\left( {\frac{{\rm{2}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{y}}}{{{\rm{25}}}}} \right)} \cdot {\rm{dy}}\\ &= \int_{\rm{0}}^{\rm{5}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy + }}\int_{\rm{5}}^{\rm{8}} {\frac{{\rm{2}}}{{\rm{5}}}} \cdot {\rm{dy - }}\int_{\rm{5}}^{\rm{8}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\int_{\rm{0}}^{\rm{5}} {\rm{y}} \cdot {\rm{dy + }}\frac{{\rm{2}}}{{\rm{5}}}\int_{\rm{5}}^{\rm{8}} {\rm{d}} {\rm{y - }}\frac{{\rm{1}}}{{{\rm{25}}}}\int_{\rm{5}}^{\rm{8}} {\rm{y}} \cdot {\rm{dy}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{0}}^{\rm{5}}{\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(y)}}_{\rm{5}}^{\rm{8}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{5}}^{\rm{8}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{5}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{0}}^{\rm{2}}}}}{{\rm{2}}}} \right){\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(8 - 5) - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{8}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{5}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}{\left( {\frac{{{\rm{25}}}}{{\rm{2}}}} \right)^{\rm{2}}}{\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(3) - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{\rm{39}}}}{{\rm{2}}}} \right)\\ &= \frac{{{\rm{25}}}}{{{\rm{50}}}}{\rm{ + }}\frac{{{\rm{60}}}}{{{\rm{50}}}}{\rm{ - }}\frac{{{\rm{39}}}}{{{\rm{50}}}}\\ &= \frac{{{\rm{46}}}}{{{\rm{50}}}}\\{\rm{P(Y}} \le 8) &= 0 {\rm{.92}}\end{aligned}\)
Therefore, the value is obtained as \({\rm{0}}{\rm{.92}}\).
Finding the Probability
(e)
The probability that the total waiting time is between \({\rm{3}}\) and \({\rm{8}}\) min is given as \({\rm{P(3}} \le {\rm{Y}} \le {\rm{8)}}\). Using the pdf \({\rm{f(y)}}\), can be written as 鈥�
\({\rm{P(3}} \le {\rm{Y}} \le {\rm{8) = P(Y}} \le {\rm{8) - P(Y}} \le {\rm{3)}}\)
Substitute the values of \({\rm{P(Y}} \le {\rm{3)}}\) and \({\rm{P(Y}} \le {\rm{8)}}\) in the above expression and solve 鈥�
\(\begin{aligned}{\rm{P(3}} \le {\rm{Y}} \le 8) &= P(Y \le {\rm{8) - P(Y}} \le {\rm{3)}}\\ &= 0 {\rm{.92 - 0}}{\rm{.18}}\\ &= 0 {\rm{.74}}\end{aligned}\)
Therefore, the value is obtained as \({\rm{0}}{\rm{.74}}\).
Finding the Probability
(f)
The probability that the total waiting time is either less than \({\rm{2}}\) min or more than \(6\) min can be written as \({\rm{P(Y}} \le {\rm{2 or 6}} \le {\rm{Y)}}\). Since the intervals \({\rm{Y}} \le {\rm{2}}\)and \({\rm{6}} \le {\rm{Y}}\)do not intersect, hence 鈥�
\({\rm{P(Y}} \le {\rm{2 or 6}} \le {\rm{Y) = P(Y}} \le {\rm{2) + P(6}} \le {\rm{Y)}}\)
Using the pdf \({\rm{f(y)}}\), \({\rm{P(Y}} \le {\rm{2)}}\) and \({\rm{P(6}} \le {\rm{Y)}}\) can be written as 鈥�
\(\begin{array}{c}{\rm{P(Y}} \le {\rm{2) = }}\int_{\rm{0}}^{\rm{2}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy}}\\{\rm{P(6}} \le {\rm{Y) = }}\int_{\rm{6}}^{{\rm{10}}} {\left( {\frac{{\rm{2}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{y}}}{{{\rm{25}}}}} \right)} \cdot {\rm{dy}}\end{array}\)
Using these in expression of \({\rm{P(Y}} \le {\rm{2 or 6}} \le {\rm{Y)}}\), it can be written 鈥�
\(\begin{aligned}{\rm{P(Y}} \le {\rm{2 or 6}} \le Y) &= \int_{\rm{0}}^{\rm{2}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy + }}\int_{\rm{6}}^{{\rm{10}}} {\left( {\frac{{\rm{2}}}{{\rm{5}}}{\rm{ - }}\frac{{\rm{y}}}{{{\rm{25}}}}} \right)} \cdot {\rm{dy}}\\ & = \int_{\rm{0}}^{\rm{2}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy + }}\int_{\rm{6}}^{{\rm{10}}} {\frac{{\rm{2}}}{{\rm{5}}}} \cdot {\rm{dy - }}\int_{\rm{6}}^{{\rm{10}}} {\frac{{\rm{y}}}{{{\rm{25}}}}} \cdot {\rm{dy}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\int_{\rm{0}}^{\rm{2}} {\rm{y}} \cdot {\rm{dy + }}\frac{{\rm{2}}}{{\rm{5}}}\int_{\rm{6}}^{{\rm{10}}} {\rm{d}} {\rm{y - }}\frac{{\rm{1}}}{{{\rm{25}}}}\int_{\rm{6}}^{{\rm{10}}} {\rm{y}} \cdot {\rm{dy}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{0}}^{\rm{2}}{\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(y)}}_{\rm{6}}^{{\rm{10}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{6}}^{{\rm{10}}}\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{{\rm{2}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{0}}^{\rm{2}}}}}{{\rm{2}}}} \right){\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(10 - 6) - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{6}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\ &= \frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{\rm{4}}}{{\rm{2}}}} \right){\rm{ + }}\frac{{\rm{2}}}{{\rm{5}}}{\rm{(4) - }}\frac{{\rm{1}}}{{{\rm{25}}}}\left( {\frac{{{\rm{64}}}}{{\rm{2}}}} \right)\\ &= \frac{{\rm{4}}}{{{\rm{50}}}}{\rm{ + }}\frac{{{\rm{80}}}}{{{\rm{50}}}}{\rm{ - }}\frac{{{\rm{64}}}}{{{\rm{50}}}} &= \frac{{{\rm{20}}}}{{{\rm{50}}}}\\{\rm{P(Y}} \le {\rm{2 or 6}} \le Y) &= 0 {\rm{.4}}\end{aligned}\)
Therefore, the value is obtained as \({\rm{0}}{\rm{.4}}\).
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