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One interesting phenomenon of bacteriuria is that there is a "turnover"; that is, if bacteriuria is measured on the same woman at two different points in time, the results are not necessarily the same. Assume that \(20 \%\) of all women who are bacteriuric at time 0 are again bacteriuric at time 1 (1 year later), whereas only 4.2\% of women who were not bacteriuric at time 0 are bacteriuric at time 1 . Let \(X\) be the random variable representing the number of bacteriuric events over the two time periods for 1 woman and still assume that the probability that a woman will be positive for bacteriuria at any one exam is \(5 \%\) What is the variance of \(X ?\)

Short Answer

Expert verified
The variance of \(X\) is approximately 0.1099.

Step by step solution

01

Define the random variable

We have defined a random variable \(X\) for the number of bacteriuric events across two time periods. A woman can have 0, 1, or 2 bacteriuric events.
02

Assign probabilities

Given the initial probabilities:- The probability of being bacteriuric at time 0, \( P(B_{0}) = 0.05 \).- The probability of being bacteriuric at time 1 given bacteriuria at time 0, \( P(B_{1} | B_{0}) = 0.20 \).- The probability of being bacteriuric at time 1 given no bacteriuria at time 0, \( P(B_{1} | eg B_{0}) = 0.042 \).
03

Calculate marginal probabilities

Calculate the marginal probability of being bacteriuric at time 1:\[ P(B_{1}) = P(B_{1} | B_{0}) \cdot P(B_{0}) + P(B_{1} | eg B_{0}) \cdot (1 - P(B_{0})) = 0.20 \cdot 0.05 + 0.042 \cdot 0.95 \]This simplifies to:\[ P(B_{1}) = 0.01 + 0.0399 = 0.0499 \] approximately 5\%.
04

Probability distribution of X

Determine the probabilities for \(X\):- \(P(X=0)\) (no events): \( (1-P(B_{0}))(1-P(B_{1}|eg B_{0})) = 0.95 \times 0.958 = 0.911 \).- \(P(X=1)\) (one event): \( P(B_{0})(1-P(B_{1}|B_{0})) + (1-P(B_{0}))P(B_{1}|eg B_{0}) = (0.05 \times 0.80) + (0.95 \times 0.042) = 0.040 + 0.0399 = 0.0799 \).- \(P(X=2)\) (two events): \( P(B_{0})(P(B_{1}|B_{0})) = 0.05 \times 0.20 = 0.01 \).
05

Calculate Expectation

The expected value of \(X\) is:\[ E(X) = 0 \cdot P(X=0) + 1 \cdot P(X=1) + 2 \cdot P(X=2) \]\[ E(X) = 0 + 0.0799 + 0.02 = 0.0999 \].
06

Calculate Variance

The variance of \(X\) is calculated using:\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]First, compute \(E(X^2)\):\[ E(X^2) = 0^2 \cdot P(X=0) + 1^2 \cdot P(X=1) + 2^2 \cdot P(X=2) \]\[ E(X^2) = 0 + 0.0799 + 0.04 = 0.1199 \]Then, substitute in the variance formula:\[ \text{Var}(X) = 0.1199 - (0.0999)^2 = 0.1199 - 0.00998001 \]\[ \text{Var}(X) = 0.10991999 \].

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Key Concepts

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

Bacteriuria
Bacteriuria refers to the presence of bacteria in the urine, which can be an indication of a urinary tract infection (UTI). For many women, the occurrence of bacteriuria can change over time, known as "turnover." This means that bacteriuria detected at one point in time might not be found at a later time. Examining bacteriuria in a statistical framework helps us understand its frequency and distribution within a population.
In the given exercise, it's noted that 20% of women who are bacteriuric at one point maintain this status a year later. On the other hand, 4.2% of women who initially test negative for bacteriuria may test positive the following year. Such data assists in modeling and predicting bacteriuric trends using statistical methods.
Random Variables
In statistics, a random variable is a variable that takes on different values randomly. In the context of bacteriuria, we can define a random variable, \( X \), to represent the number of bacteriuric events over two time periods for a single woman. This means \( X \) can take values of 0, 1, or 2, representing zero, one, or two bacteriuric events observed.
Random variables are crucial in statistical models because they simplify the process of describing random phenomena through numerical values. By using \( X \), we can quantitatively analyze and make probabilistic predictions about future events related to bacteriuria.
Probability Distribution
Probability distribution describes how likely each possible value of a random variable is. For \( X \), the number of bacteriuric events, the distribution involves calculating the probabilities of 0, 1, or 2 events.
Given the exercise data:
  • \( P(X = 0) \) – no events occur, which means neither initial nor subsequent bacteriuria is detected.
  • \( P(X = 1) \) – one event occurs, either initial or subsequent bacteriuria is detected.
  • \( P(X = 2) \) – both events occur, bacteriuria is detected at both times.
Probabilities are calculated using known conditional probabilities. This distribution is key to understanding the gist of bacteriuria turnover rates.
Variance Calculation
Variance is a measure of how much a set of numbers (or a random variable) is spread out. It tells us about the variability of the random variable outcomes. For \( X \), variance helps us understand the variance in bacteriuria events.
To calculate variance, we first determine the expected value \( E(X) \) and \( E(X^2) \) (which is the expectation of \( X \) squared). The formula for variance becomes
    \[ \text{Var}(X) = E(X^2) - (E(X))^2 \]
Calculating \( \text{Var}(X) \) provides insight into how much the number of bacteriuria events deviates from the expected average. A high variance would suggest significant variability in bacteriuric events among women, while a low variance would indicate more consistency.

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