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Chapter 12: Problem 34

In a certain cell population, cells divide every 10 days, and the age of a cell selected at random is a random variable \(X\) with the density function \(f(x)=2 k e^{-k x}, 0 \leq x \leq 10, k=(\ln 2) / 10\). A random variable \(X\) has a density function \(f(x)=\frac{2}{3} x\) on \(1 \leq x \leq 2\). Find \(a\) such that \(\operatorname{Pr}(a \leq X)=\frac{1}{3}\).

Short Answer

Expert verified
The value of \(a\) is \(\sqrt{3}\).

Step by step solution

01

Understand the Given Density Function

The density function for the random variable \(X\) is given by \(f(x) = \frac{2}{3} x\) on the interval \(1 \leq x \leq 2\). This means that \(f(x)\) is defined and applicable only between the values 1 and 2.
02

Set Up the Probability Expression

To find the value of \(a\) such that \(\operatorname{Pr}(a \leq X) = \frac{1}{3}\), set up the integral of the density function \(f(x)\) from \(a\) to 2, equating it to \(\frac{1}{3}\):\[\int_{a}^{2} \frac{2}{3} x \, dx = \frac{1}{3}.\]
03

Integrate the Density Function

Integrate \(\frac{2}{3} x\) with respect to \(x\):\[\int_{a}^{2} \frac{2}{3} x \, dx = \left[ \frac{2}{3} \cdot \frac{x^2}{2} \right]_{a}^{2} = \left[ \frac{1}{3} x^2 \right]_{a}^{2} \]
04

Solve for \(a\)

Set the expression equal to \(\frac{1}{3}\) and solve for \(a\):\[\frac{4}{3} - \frac{1}{3} a^2 = \frac{1}{3}\]Subtract \(\frac{1}{3}\) from both sides:\[\frac{4}{3} - \frac{1}{3} = \frac{1}{3} a^2\]Simplify:\[\frac{3}{3} = \frac{1}{3} a^2\]\[a^2 = 3\]Take the square root of both sides to find \(a\):\[a = \sqrt{3}\]

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Key Concepts

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

Random Variable
In probability and statistics, a random variable is a variable that takes on different values randomly. These values depend on the outcomes of a random phenomenon.
For example, consider the age of a cell in a population where cells divide every 10 days. Here, the age of a randomly selected cell is a random variable. It's denoted by a symbol, commonly represented as \(X\).
A random variable can be discrete or continuous. A discrete random variable has a countable number of possible values, such as the number of heads in a series of coin tosses. A continuous random variable, on the other hand, has an infinite number of possible values, typically within some interval.
  • In our exercise, the random variable \(X\) represents the age of a cell, which is continuous within the interval \[0, 10\].
  • The probability density function (PDF) describes the likelihood of different outcomes for a continuous random variable.
  • In this case, the PDF is given as \[ f(x) = 2ke^{-kx} \], where \( k = \frac{\text{ln} 2}{10}.\)
Integral of a Function
The integral of a function is a key concept in calculus. It helps in finding areas, volumes, central points, and many other meaningful values.
When dealing with probability density functions (PDFs), integrating the function over a certain range gives the probability that the random variable falls within that range.
  • For continuous random variables, the integral of the PDF over all possible values must be equal to 1, representing a total probability of 100%.
  • In our given exercise, we need to find \(a\), such that the probability \(\text{Pr}(a \leq X) = \frac{1}{3}.\)
To solve this, set up the integral of the PDF from \(a\) to 2 and equate it to \(\frac{1}{3}:\)
\[\text{\int_{a}^{2} \frac{2}{3} x \ dx = \frac{1}{3}} \]
Solving Equations
Solving equations is a fundamental skill in algebra and calculus. It involves finding the values that satisfy the given mathematical expression.
In our problem, we set up an integral equation to find the value of \(a\) for a specific probability. The steps to solve such an equation typically involve:
  • Setting up the integral equation with the PDF.
  • Integrating the function with respect to the variable.
  • Solving the resulting algebraic equation for the variable of interest.
For the given interval, the steps are:
Perform the integration:
\[\text{\int_{a}^{2} \frac{2}{3} x \ dx = \left[ \frac{1}{3} x^2 \right]_{a}^{2}} \]
Set it equal to \(\frac{1}{3}\):
\[\text{\frac{4}{3} - \frac{1}{3} a^2 = \frac{1}{3}} \]
Simplify and solve for \(a\):
\[\text{\frac{4}{3} - \frac{1}{3} = \frac{1}{3} a^2} \]
  • Subtract \(\frac{1}{3}\) from both sides, simplify to get \(a^2 = 3\)
  • Take the square root of both sides, yielding \(a = \sqrt{3}.\)
Hence, \(a = \sqrt{3}\) is the solution ensuring the probability \(\text{Pr}(a \leq X) = \frac{1}{3}.\)

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Most popular questions from this chapter

Time of a Commute A student with an eight o'clock class at the University of Maryland commutes to school by car. She has discovered that along each of two possible routes her traveling time to school (including the time to get to class) is approximately a normal random variable. If she uses the Capital Beltway for most of her trip, \(\mu=25\) minutes and \(\sigma=5\) minutes. If she drives a longer route over local city streets, \(\mu=28\) minutes and \(\sigma=3\) minutes. Which route should the student take if she leaves home at 7:30 A.M.? (Assume that the best route is one that minimizes the probability of being late to class.)

Find the expected value and variance for each random variable whose probability density function is given. When computing the variance, use formula (5). \(f(x)=\frac{8}{9} x, 0 \leq x \leq \frac{3}{2}\)

Find the expected value and variance for each random variable whose probability density function is given. When computing the variance, use formula (5). \(f(x)=\frac{1}{18} x, 0 \leq x \leq 6\)

Compute the cumulative distribution function corresponding to the density function \(f(x)=\frac{1}{2}(3-x)\), \(1 \leq x \leq 3 .\)

Maximum Likelihood Exercises 19 and 20 illustrate a technique from statistics (called the method of maximum likelihood) that estimates a parameter for a probability distribution. In a production process, a box of fuses is examined and found to contain two defective fuses. Suppose that the probability of having two defective fuses in a box selected at random is \(\left(\lambda^{2} / 2\right) e^{-\lambda}\) for some \(\lambda\). Take first and second derivatives to determine the value of \(\lambda\) for which the probability has its maximum value.
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