Problem 64 - Calculus for Biology and Medicine

Question

The hazard-rate function of an organism is given by $$ \lambda(x)=0.1+0.5 e^{0.02 x}, \quad x \geq 0 $$ where \(x\) is measured in days. (a) What is the probability that the organism will live less than 10 days? (b) What is the probability that the organism will live for another five days, given that it survived the first five days?

Step-by-Step Solution

Answer

(a) Calculate and find \( P(X < 10) \) by taking \( 1 - S(10) \). (b) Use: \( \frac{S(10)}{S(5)} \) and compute using \( S(x) \) values.

Step 1: Understand the Hazard Rate Function

The hazard-rate function \( \lambda(x) \) represents the instantaneous rate of failure at time \( x \). For this exercise, it is given by \( \lambda(x) = 0.1 + 0.5 e^{0.02 x} \). This function will help us determine the survival function and the required probabilities.

Step 2: Define the Survival Function

The survival function \( S(x) \) is related to the hazard function by the equation \( S(x) = e^{- ext{Integral of } \lambda(x)} \). Thus, \( S(x) = e^{ -\int_{0}^{x} (0.1 + 0.5 e^{0.02 t}) \, dt } \).

Step 3: Compute the Integral for Survival Function

Calculate \( \int_{0}^{x} (0.1 + 0.5 e^{0.02 t}) \, dt \). - For \( 0.1 \), the integral is \( 0.1x \).- For \( 0.5 e^{0.02 t} \), the integral is \( 0.5 \times \frac{1}{0.02} e^{0.02 t} \bigg|_{0}^{x} = 25 (e^{0.02 x} - 1) \).Thus, \( \int_{0}^{x} (0.1 + 0.5 e^{0.02 t}) \, dt = 0.1x + 25(e^{0.02 x} - 1) \).

Step 4: Derive the Expression for Survival Function

Using the integral from Step 3, the survival function is given by:\[ S(x) = e^{- \left( 0.1x + 25(e^{0.02 x} - 1) \right) } = e^{ -0.1x - 25(e^{0.02 x} - 1) } \].

Step 5: Calculate Probability for Part (a)

To find the probability that the organism will live less than 10 days, calculate:\( P(X < 10) = 1 - S(10) \).Substitute into \( S(x) \):\[ S(10) = e^{ -0.1 \times 10 - 25(e^{0.02 \times 10} - 1) } \].Using a calculator, find:\[ S(10) \approx e^{ -1 - 25(e^{0.2} - 1) } \] and then compute \( 1 - S(10) \).

Step 6: Calculate Probability for Part (b)

To find the probability the organism survives another 5 days, given it survived the first 5, use:\[ P(X > 10 | X > 5) = \frac{S(10)}{S(5)} \].Substitute values:\[ S(5) = e^{ -0.1 \times 5 - 25(e^{0.02 \times 5} - 1) } \].Compute \( S(10) / S(5) \) using values from Step 5 and a calculator for precision.

Key Concepts

Survival Function

A survival function, often denoted as \( S(x) \), represents the probability that an organism or system continues to operate without failure up until a certain time \( x \). In simpler terms, it tells us the likelihood that the organism is still "alive" after \( x \) days.

The relationship between the survival function and the hazard rate function \( \lambda(x) \) is central to understanding reliability and survival analysis. The survival function is calculated using the formula:

• \( S(x) = e^{-\int_{0}^{x} \lambda(t) \, dt} \)

This equation uses the integral of the hazard function over time to calculate the exponent for the exponential function. The outcome is a decreasing function indicating decreasing probability of survival with increasing time.

In the given problem, the survival function for the organism is described by this formula:

• \( S(x) = e^{- (0.1x + 25(e^{0.02x} - 1))} \)

This expression is derived from the function \( \lambda(x) = 0.1 + 0.5 e^{0.02 x} \). Understanding this linkage helps in assessing the probability that an organism survives beyond certain time frames.

Probability Calculation

Probability calculation in survival analysis is used to find the likelihood of certain time-bound outcomes concerning the lifespan of organisms. Typically, such calculations involve the survival function \( S(x) \) and its complement \( P(X < x) = 1 - S(x) \), which gives the probability of failing before a particular time.

In our exercise, two probabilities are of interest:

• The probability that an organism dies before 10 days: The calculation involves first determining \( S(10) \) using the survival function and then finding \( 1 - S(10) \). This gives the probability that the organism does not survive up to 10 days.
• The conditional probability that the organism lives for an additional 5 days, given it has already survived 5 days: This is found using \( P(X > 10 | X > 5) = \frac{S(10)}{S(5)} \). This ratio of survival probabilities reflects how much longer an organism is likely to live given its current state.

By plugging in the appropriate calculations into these formulas, you can accurately estimate these probabilities, providing insights into the survival dynamics of the organism.

Integral Calculus

Integral calculus is vital in analyzing survival data as it helps us determine the survival function from the hazard rate. It involves calculating the area under the curve for the hazard function \( \lambda(x) \) from time \( 0 \) to time \( x \). This integral gives us the exponent in the survival function formula:

\[ \int_{0}^{x} \lambda(t) \, dt \]

For our exercise, this requires integrating:

• \( 0.1 \), resulting in \( 0.1x \)
• \( 0.5 e^{0.02 t} \), leading to \( 25(e^{0.02x} - 1) \)

Thus, the complete integral becomes \( 0.1x + 25(e^{0.02x} - 1) \). This simplifies our survival function to \( S(x) = e^{- (0.1x + 25(e^{0.02x} - 1))} \).

Grasping these integrals is a key step as they detail how quickly the related risk grows with time, which influences how drastic initial survival assumptions might change in light of such mathematical results.