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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?

Short Answer

Expert verified

(a) The value of\({\rm{k}}\)is obtained as\({\rm{k = }}\frac{{\rm{3}}}{{\rm{8}}}\).

(b) The probability that the lecture ends within\({\rm{1}}\)min of the end of the hour is\({\rm{0}}{\rm{.125}}\).

(c) The probability that the lecture continues beyond the hour for between\({\rm{60}}\)and\({\rm{90}}\)sec is\({\rm{0}}{\rm{.2969}}\).

(d) The probability that the lecture continues for at least \({\rm{90}}\) sec beyond the end of the hour is \({\rm{0}}{\rm{.5781}}\).

Step by step solution

01

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.

02

The value for \({\rm{k}}\)

(a)

Since\({\rm{f(x)}}\)is a legitimate pdf, hence it must satisfy following condition 鈥

\(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(x)}} \cdot {\rm{dx = 1}}} \)

For the given pdf 鈥

\(\begin{aligned}{\rm{k}}\int_{{\rm{ - }}\infty }^\infty {\rm{f}} {\rm{(x)}} \cdot dx &= \int_{\rm{0}}^{\rm{2}} {\rm{k}} {{\rm{x}}^{\rm{2}}} \cdot {\rm{dx = 1}}\\\left( {\frac{{{\rm{k}}{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{0}}^{\rm{2}} &= 1\\\frac{{{\rm{k(2}}{{\rm{)}}^{\rm{3}}}}}{{\rm{3}}}- 0 &= 1\\ k &= \frac{{\rm{3}}}{{\rm{8}}}\end{aligned}\)

Hence the pdf can finally be written as 鈥

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\frac{{{\rm{3}}{{\rm{x}}^{\rm{2}}}}}{{\rm{8}}}}&{{\rm{1}} \le {\rm{X}} \le {\rm{2}}}\\{{\rm{0}}\;\;\;}&{{\rm{otherwise}}}\end{array}} \right.\)

Conditions satisfied by a pdf: For a pdf to be a legitimate pdf, it must satisfy the following two conditions 鈥

  1. \({\rm{f(x)}} \ge {\rm{0}}\)for all\({\rm{x}}\)
  2. \(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(x)}} \cdot {\rm{dx = 1}}} \)

Therefore, the value is obtained as \({\rm{k = }}\frac{{\rm{3}}}{{\rm{8}}}\).

03

Finding the probability

(b)

The probabilitythat lecture ends within \({\rm{1}}\) minute of the end of the hour is denoted as \({\rm{P(X < 1)}}\).

\(\begin{array}{c}{\rm{P(X < 1) = }}\int_{\rm{0}}^{\rm{1}} {\frac{{{\rm{3}}{{\rm{x}}^{\rm{2}}}}}{{\rm{8}}}} \cdot {\rm{dx}}\\{\rm{ = }}\left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{8}}}} \right)_{\rm{0}}^{\rm{1}}\\{\rm{ = }}\left( {\frac{{\rm{1}}}{{\rm{8}}}{\rm{ - 0}}} \right)\\{\rm{ = 0}}{\rm{.125}}\end{array}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.125}}\).

04

Finding the probability

(c)

The probability that lecture continues beyond the hour for between \({\rm{60}}\) and \({\rm{90}}\) seconds is denoted as \({\rm{P(X < 1}}{\rm{.5)}}\).

\(\begin{aligned}{\rm{P(1 < X < 1}}.5) &= \int_{\rm{1}}^{{\rm{1}}{\rm{.5}}} {\frac{{{\rm{3}}{{\rm{x}}^{\rm{2}}}}}{{\rm{8}}}} \cdot {\rm{dx}}\\ &= \left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{8}}}} \right)_{\rm{1}}^{{\rm{1}}{\rm{.5}}}\\ &= \left( {\frac{{{{{\rm{(1}}{\rm{.5)}}}^{\rm{3}}}}}{{\rm{8}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{8}}}} \right)\\ &= 0 {\rm{.2969}}\end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.2969}}\).

05

Finding the probability

(d)

The probability that lecture continues at least \({\rm{90}}\) seconds beyond the end of the hour is denoted as \({\rm{P(X}} \ge {\rm{1}}{\rm{.5)}}\).

\(\begin{aligned}{\rm{P(X}} \ge {\rm{1}} .5) &= \int_{1.5}^2 {\frac{{{\rm{3}}{{\rm{x}}^{\rm{2}}}}}{{\rm{8}}}} \cdot {\rm{dx}}\\ &= \left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{8}}}} \right)_{1.5}^2\\ &= \left( {\frac{{{{(2)}^3}}}{{\rm{8}}}{\rm{ - }}\frac{{{{(1.5)}^3}}}{{\rm{8}}}} \right)\\ & = 0{\rm{.5781}}\end{aligned}\)

Therefore, the value is obtained as \({\rm{0}}{\rm{.5781}}\).

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