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Chapter 3: Problem 203

A type of bacteria cell divides at a constant rate \(\lambda\) over time. (That is, the probability that a cell divides in a small interval of time \(t\) is approximately \(\lambda t\).) Given that a population starts out at time zero with \(k\) cells of this bacteria and that cell divisions are independent of one another, the size of the population at time \(t, Y(t),\) has the probability distribution $$P[Y(t)=n]=\left(\begin{array}{l}n-1 \\\k-1\end{array}\right) e^{-\lambda k t}\left(1-e^{-\lambda t}\right)^{n-k}, \quad n=k, k+1, \ldots$$ b. If, for a type of bacteria cell, \(\lambda=.1\) per second and the population starts out with two cells at time zero, find the expected value and variance of the population after five seconds.

Short Answer

Expert verified
Expected value is approximately 3.2974; variance is approximately 1.2974.

Step by step solution

01

Define the given parameters

We are given that the division rate \( \lambda = 0.1 \) per second, initial number of cells \( k = 2 \), and we are interested in the time \( t = 5 \) seconds.
02

Identify the distribution involved

The random variable \( Y(t) \) follows a negative binomial distribution, which in this context, due to the properties of the Poisson process with constant rate and independent events, can be treated as a Poisson distribution with parameter \( m(t) = \lambda t \).
03

Calculate the expected value

For a population following a Poisson process, the expected number of divisions after time \( t \) is given by \( E[Y(t)] = k e^{\lambda t} \). Substituting the known values: \( E[Y(5)] = 2 e^{0.1 \times 5} \), we calculate this to find the expected value.
04

Calculate \( e^{0.5} \)

Evaluate \( e^{0.5} \), where \( \lambda t = 0.5 \). \( e^{0.5} \approx 1.6487 \).
05

Compute the expected value

Substitute the computed values in the expression: \( E[Y(5)] = 2 \times 1.6487 \approx 3.2974 \). Thus, the expected size of the population after five seconds is approximately 3.2974.
06

Determine the variance

For a Poisson process, the variance of the population size \( Y(t) \) after time \( t \) is also given by \( \text{Var}[Y(t)] = k e^{\lambda t} - k \) because the initial population adds \( k \) to the variance. So, \( \text{Var}[Y(5)] = 2 (e^{0.5} - 1) \).
07

Calculate the variance

Compute the variance: \( \text{Var}[Y(5)] = 2 (1.6487 - 1) = 2 \times 0.6487 = 1.2974 \). Thus, the variance of the population size after five seconds is approximately 1.2974.

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Key Concepts

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

Negative Binomial Distribution
When studying processes like bacteria cell division, it's important to understand certain probability distributions. The **Negative Binomial Distribution** is one such statistical tool. This distribution is particularly useful when we're dealing with the number of successes in a series of independent experiments. - The definition involves parameters such as the number of successes we want and the probability of each success. - Specifically, it shows how many trials are needed to achieve a predetermined number of successes. In the context of bacteria, imagine you want to know how many attempts it takes for a certain number of bacteria to divide successfully. While complex, this model helps in predicting the distribution of such events over time.
Expected Value
The **Expected Value** is a crucial concept in probability and statistics. It provides a measure of the "central" or "average" outcome expected from a random variable. For our bacteria cell division, calculating expected value helps predict the average size of a population after a defined time is elapsed, based on the rate of division. To find the expected value: - Identify the initial number of cells or trials. - Use the formula involving the known division rate and time. In our example, after 5 seconds, we expected the bacteria population to size up to approximately 3.2974 cells. This calculation is crucial for researchers to gauge growth efficiency or make forecasts on population dynamics.
Variance
**Variance** measures how much the outcomes of a random variable like bacteria cell division differ from the mean or expected value. It shows the variability or spread in the dataset, giving us an idea about the distribution consistency. To compute the variance for our Poisson process related to bacteria: - Use the variance formula adapted for initial populations. It typically involves factors like the number of initial cells and the rate of nucleus division minus adjustments for starting conditions. For our case, the variance was calculated to be approximately 1.2974, giving insights into how variable future cell populations might be after the same time duration.
Bacteria Cell Division
**Bacteria cell division** is a fascinating natural process and a foundation for understanding microbial population dynamics. In simple terms, bacteria reproduce through a method called binary fission, where one cell splits into two. - This process is affected by various factors such as nutrient availability, temperature, and innate biological rates. - Researchers often model cell division rates using mathematical frameworks to understand growth under different conditions. In studies, knowing the division rate of bacteria is crucial for applications in health, where controlling bacterial populations can impact infection management, or in industry, where specific growth rates are needed for production.
Probability Distribution
A **Probability Distribution** is a statistical function that defines all the possible outcomes for a random variable and how probabilities are assigned to those outcomes. It gives a complete view of how values distribute themselves in a specific problem. - Common probability distributions include Normal, Poisson, and the Negative Binomial. - For problems like bacteria cell division, suited distributions like Poisson are used, since events happen at a constant average rate independently. Understanding probability distributions helps in predicting events, assessing the variance of outcomes, and effectively planning for scenarios based on probabilistic expectations.

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Most popular questions from this chapter

The probability of a customer arrival at a grocery service counter in any one second is equal to. \(1 .\) Assume that customers arrive in a random stream and hence that an arrival in any one second is independent of all others. Find the probability that the first arrival a. will occur during the third one-second interval. b. will not occur until at least the third one-second interval.

Let \(Y\) have a hypergeometric distribution $$p(y)=\frac{\left(\begin{array}{l} r \\ y \end{array}\right)\left(\begin{array}{l} N-r \\ n-y \end{array}\right)}{\left(\begin{array}{l} N \\ n \end{array}\right)} \quad y=0,1,2, \ldots, n$$ a. Show that $$P(Y=n)=p(n)=\left(\frac{r}{N}\right)\left(\frac{r-1}{N-1}\right)\left(\frac{r-2}{N-2}\right) \cdots\left(\frac{r-n+1}{N-n+1}\right)$$ b. Write \(p(y)\) as \(p(y | r) .\) Show that if \(r_{1}\frac{p\left(y+1 | r_{1}\right)}{p\left(y+1 | r_{2}\right)}$$ c. Apply the binomial expansion to each factor in the following equation: $$(1+a)^{N_{1}}(1+a)^{N_{2}}=(1+a)^{N_{1}+N_{2}}$$ Now compare the coefficients of \(a^{n}\) on both sides to prove that $$\left(\begin{array}{c} N_{1} \\ 0 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n \end{array}\right)+\left(\begin{array}{c} N_{1} \\ 1 \end{array}\right)\left(\begin{array}{c} N_{2} \\ n-1 \end{array}\right)+\cdots+\left(\begin{array}{c} N_{1} \\ n \end{array}\right)\left(\begin{array}{c} N_{2} \\ 0 \end{array}\right)=\left(\begin{array}{c} N_{1}+N_{2} \\ n \end{array}\right)$$ d. Using the result of part ( \(\underline{\mathrm{c}}\) ), conclude that $$\sum_{y=0}^{n} p(y)=1$$

The number of bacteria colonies of a certain type in samples of polluted water has a Poisson distribution with a mean of 2 per cubic centimeter \(\left(\mathrm{cm}^{3}\right)\) a. If four \(1-\mathrm{cm}^{3}\) samples are independently selected from this water, find the probability that at least one sample will contain one or more bacteria colonies. b. How many \(1-\mathrm{cm}^{3}\) samples should be selected in order to have a probability of approximately .95 of seeing at least one bacteria colony?

Let \(Y\) have a Poisson distribution with mean \(\lambda\). Find \(E[Y(Y-1)]\) and then use this to show that \(V(Y)=\lambda\).

Suppose that \(Y\) is a discrete random variable with mean \(\mu\) and variance \(\sigma^{2}\) and let \(W=2 Y\) a. Do you expect the mean of \(W\) to be larger than, smaller than, or equal to \(\mu=E(Y) ?\) Why? b. Use Theorem 3.4 to express \(E(W)=E(2 Y)\) in terms of \(\mu=E(Y) .\) Does this result agree with your answer to part (a)? c. Recalling that the variance is a measure of spread or dispersion, do you expect the variance of \(W\) to be larger than, smaller than, or equal to \(\sigma^{2}=V(Y) ?\) Why? d. Use Definition 3.5 and the result in part (b) to show that $$V(W)=E\left\\{[W-E(W)]^{2}\right\\}=E\left[4(Y \mu)^{2}\right]=4 \sigma^{2}$$; that is, \(W=2 Y\) has variance four times that of \(Y\).
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