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Chapter 12: Problem 32

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 \(f(x)=2 k e^{-k x}, 0 \leq x \leq 10, k=(\ln 2) / 10\). Upon examination of a slide, \(10 \%\) of the cells are found to be undergoing mitosis (a change in the cell leading to division). Compute the length of time required for mitosis; that is, find the number \(M\) such that $$ \int_{10-M}^{10} 2 k e^{-k x} d x=.10 $$

Short Answer

Expert verified
M=10-\frac{10 \ln 0.55}{ \ln 2}.

Step by step solution

01

Identify the Given Parameters

We have a cell population where cells divide every 10 days. The density function is given by \( f(x) = 2 k e^{-k x} \) for \( 0 \leq x \leq 10 \) with \( k = \frac{\ln 2}{10} \).
02

Set Up the Integral

We need to find the value of \( M \) such that the integral of the density function from \( 10 - M \) to \( 10 \) equals 0.10. This is written as: \[ \int_{10-M}^{10} 2 k e^{-k x} d x = 0.10 \]
03

Solve the Integral

First, evaluate the integral: \[ \int_{10-M}^{10} 2 k e^{-k x} d x = 2k \left[-\frac{1}{k} e^{-k x}\right]_{10-M}^{10} \] Simplify the expression: \[ = 2 \left[- e^{-k x}\right]_{10-M}^{10} \] \[ = 2 \left[ -e^{-k \cdot 10} + e^{-k (10-M)} \right] \] Substitute \( k = \frac{\ln 2}{10} \): \[ = 2 \left[ -e^{-\frac{10 \ln 2}{10}} + e^{-\frac{(10-M) \ln 2}{10}} \right] \]\[ = 2 \left[ -e^{-\ln 2} + e^{-\frac{(10-M) \ln 2}{10}} \right] \] Since \( e^{\ln a} = a \) and \( e^{-\ln 2} = \frac{1}{2} \), continue simplifying: \[ = 2 \left[ -\frac{1}{2} + e^{-\frac{\ln 2 (10-M)}{10}} \right] \] \[ = 2 \left[ -\frac{1}{2} + e^{-\frac{\ln 2 (10-M)}{10}} \right] \] Since \( e^{-\ln 2} = \frac{1}{2} \), we change the expression:
04

Set the Integral Equal to 0.10

We know that \[ 2 \left[ -\frac{1}{2} + e^{-\frac{\ln 2 (10-M)}{10}} \right] = 0.10 \]Solve for \( M \):\[ -1 + 2e^{-\frac{\ln 2}{10} (10-M)} = 0.1 \]\[ 2e^{-\frac{\ln 2}{10} (10-M)} = 1.1 \]\[ e^{-\frac{\ln 2}{10} (10-M)}=0.55 \] Take the natural log: \[ -\frac{\ln 2}{10} (10-M)=\ln 0.55 \]\[ 10-M= \frac{10 \ln 0.55}{\ln 2} \]\[ M= 10 -\frac{10 \ln 0.55}{ \ln 2} \]

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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 based on the outcome of a random phenomenon. In the given exercise, the age of a cell selected at random is regarded as a random variable denoted as \(X\). This means the age is not fixed but varies depending on which cell is chosen from the population.
Understanding that cell age is a random variable helps us grasp that certain probabilities can be assigned to the different possible ages.
For example, if a cell divides every 10 days, the age falls within the interval \([0, 10]\) days with a specified probability. Values of the random variable can then be analyzed using various statistical tools and functions, such as the density function.
Density Function
A density function is crucial in probability theory. It describes the likelihood of a random variable taking on a range of values. For the given exercise, the density function of cellular age is:
\[ f(x) = 2 k e^{-k x} \text{ for } 0 \leq x \leq 10 \text{ with } k = \frac{\ln 2}{10} \]
This function tells us how the probability is distributed over the interval \([0, 10]\).
In simpler terms, the density function helps in determining the probability for any given cell's age and is tailored to show the function's behavior and key parameters, such as \(k\). Understanding the density function shape and properties will facilitate the computation of probabilities for specific cell ages.
Integral Calculus
Integral calculus is a critical tool for solving problems related to continuous probability distributions. In this case, we use integration to find the cumulative probability over an interval. The exercise requires us to compute the length of time required for mitosis by setting up and solving the following integral:
\[ \int_{10-M}^{10} 2k e^{-k x} dx = 0.10 \]
Integration aids in determining the area under the density function curve between \(10-M\) and \(10\), representing the probability that a cell is in the mitosis phase.
The integral's solution, through techniques of substitution and evaluating limits, helps identify the length of the mitosis duration.
Mitosis Duration
The question seeks to calculate the duration of mitosis, requiring precise understanding of the bounds within which cells undergo this process. Mitosis represents a change leading to cell division. In terms of probability, finding the mitosis duration entails solving for \( M \) such that:
\[ 2 \left[ -\frac{1}{2} + e^{-\frac{\ln 2 (10-M)}{10}} \right] = 0.10 \]
The calculations involve determining the value of \(M\) that delineates the time frame ending at day 10. With the cells dividing every 10 days, exactly 10% of cells are undergoing mitosis.
We equate the integral result to 0.10 and solve algebraically, taking the natural logarithm when necessary, leading to the final solution:
\[ M = 10 - \frac{10 \ln 0.55}{ \ln 2} \]This final value of \(M\) quantifies the duration during which mitosis occurs within each 10-day cycle for the cell population.

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Most popular questions from this chapter

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