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Chapter 4: Problem 5

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within \(2 \mathrm{~min}\) after the hour. Let \(X=\) the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of \(X\) is $$ f(x)=\left\\{\begin{array}{cc} k x^{2} & 0 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Find the value of \(k\). [Hint: Total area under the graph of \(f(x)\) is 1.] b. What is the probability that the lecture ends within \(1 \mathrm{~min}\) of the end of the hour? c. What is the probability that the lecture continues beyond the hour for between 60 and \(90 \mathrm{~s}\) ? d. What is the probability that the lecture continues for at least \(90 \mathrm{~s}\) beyond the end of the hour?

Short Answer

Expert verified
a. \(k = \frac{3}{8}\); b. Probability is \(\frac{1}{8}\); c. Probability is \(0.2975\); d. Probability is \(0.5781\).

Step by step solution

01

Determine the Value of k

Since the total area under the probability density function (pdf) must equal 1, we set up the integral of the pdf over its range and solve for the constant \(k\):\[\int_{0}^{2} kx^2 \, dx = 1\]First, compute the indefinite integral:\[\int kx^2 \, dx = \frac{kx^3}{3}\]Now, evaluate from 0 to 2:\[\left.\frac{kx^3}{3}\right|_0^2 = \frac{k(2)^3}{3} - \frac{k(0)^3}{3} = \frac{8k}{3}\]Set this equal to 1:\[\frac{8k}{3} = 1 \k = \frac{3}{8}\]
02

Probability Lecture Ends Within 1 Minute

We calculate the probability that the lecture ends within 1 minute after the hour by evaluating the cumulative distribution function from 0 to 1:\[P(0 \leq X \leq 1) = \int_{0}^{1} \frac{3}{8} x^2 \, dx\]Compute the integral:\[\frac{3}{8} \int_{0}^{1} x^2 \, dx = \frac{3}{8} \left[ \frac{x^3}{3} \right]_0^1 = \frac{3}{8} \cdot \frac{1}{3} = \frac{1}{8}\]
03

Probability Lecture Continues Beyond 60 to 90 Seconds

Find the probability that the lecture ends between 60 seconds (1 minute) and 90 seconds (1.5 minutes):\[P(1 \leq X \leq 1.5) = \int_{1}^{1.5} \frac{3}{8} x^2 \, dx\]Compute the integral:\[\frac{3}{8} \int_{1}^{1.5} x^2 \, dx = \frac{3}{8} \left[ \frac{x^3}{3} \right]_1^{1.5}= \frac{3}{8} \left( \frac{(1.5)^3 - 1^3}{3} \right)\]\[= \frac{3}{8} \cdot \frac{3.375 - 1}{3} = \frac{3}{8} \cdot 0.7917 \approx 0.2975\]
04

Probability of Lecture Continuing at Least 90 Seconds

Find the probability that the lecture continues for at least 90 seconds:\[P(X \geq 1.5) = \int_{1.5}^{2} \frac{3}{8} x^2 \, dx\]Compute the integral:\[\frac{3}{8} \int_{1.5}^{2} x^2 \, dx = \frac{3}{8} \left[ \frac{x^3}{3} \right]_{1.5}^{2} = \frac{3}{8} \left( \frac{8 - 3.375}{3} \right)\]\[= \frac{3}{8} \cdot 1.5417 \approx 0.5781\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function
A Probability Density Function (pdf) describes the likelihood of a continuous random variable occurring within a particular range of values. In simpler terms, it is a function that helps determine probabilities in continuous data.
Unlike discrete probabilities, which deal with distinct outcomes, pdfs apply to outcomes within an interval. The essential property of a pdf is that the integral (area under the curve) over its entire range is equal to 1, signifying the certainty that some value within the distribution will occur.
For instance, in the problem above, the given pdf is defined as:
  • \( f(x) = k x^2 \) for \( 0 \leq x \leq 2 \)
  • \( f(x) = 0 \) otherwise.
This means that the only values with a non-zero probability occur between 0 and 2 minutes past the hour, aligning with the context that lectures end within this timeframe.
Continuous Random Variable
A Continuous Random Variable represents variables that can take an infinite number of values between any two given points. Think of it as a variable that smoothly transitions across an interval.
In our case, the time at which the lecture ends is a continuous random variable, denoted as \(X\). It can realistically take any value from 0 to 2 minutes beyond the hour, since time, in reality, can be infinitely divisible into milliseconds and beyond.
This type of variable is measured, not counted, which is why pdfs are used. They give the density of probabilities at a point by integrating over intervals of interest, rather than providing exact probabilities at specific points as with discrete variables.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) connects closely with pdfs by showing the probability that a continuous random variable, \(X\), takes on a value less than or equal to a specific value.
Essentially, the CDF is obtained by integrating the pdf from the lower possible value up to a point of interest. It aggregates probabilities incrementally from the start of the distribution range. The CDF is non-decreasing and always ranges between 0 (certainty of X being less than the smallest value of X) and 1 (certainty that X is less than or equal to the largest value of X).
In the exercise, finding probabilities like the lecture ending within 1 minute or continuing beyond certain points requires integrating the pdf over corresponding intervals, yielding the cumulative probabilities that form the CDF.
Integration in Probability
Integration is a central concept in probability, especially when dealing with continuous random variables. It allows us to find areas under the pdf curve, which correspond to the probabilities we seek.
To determine the value of \(k\) in the exercise, integration ensures the total area under the pdf is 1. Solving: \[ \int_{0}^{2} kx^{2} \, dx = 1 \]yields \( k = 3/8 \), ensuring the pdf is a valid probability distribution.
Additionally, integration is utilized to evaluate specific probabilities within given ranges, such as the lecture ending within 1 minute or continuing for over 90 seconds. These calculations involve integrating the pdf over specified intervals, thus summing up the densities to yield probabilities.
  • It gives us the cumulative probabilities needed for CDFs.
  • Is the tool that turns a pdf into actionable data, helping answer practical questions surrounding the random variable.
Understanding integration in probability helps unlock the full capabilities of pdfs and CDFs in providing meaningful insights into continuous random variables.

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