Question:medium
The function, \( r(t) \), signifies the likelihood that a component has survived up to time \( t \) and fails in the next instant of time.
\[ r(t) = f(t) / (1 - F(t)) \]
where \( f(t) \) denotes the probability density function for time to failure with corresponding cumulative distribution function given by,
\[ F(t) = 1 - e^{-at^b}, \quad a > 0, b > 0 \]
Which ONE of the following represents \( r(t) \)?
The function, \( r(t) \), signifies the likelihood that a component has survived up to time \( t \) and fails in the next instant of time.
\[ r(t) = f(t) / (1 - F(t)) \]
where \( f(t) \) denotes the probability density function for time to failure with corresponding cumulative distribution function given by,
\[ F(t) = 1 - e^{-at^b}, \quad a > 0, b > 0 \]
Which ONE of the following represents \( r(t) \)?
\[ r(t) = f(t) / (1 - F(t)) \]
where \( f(t) \) denotes the probability density function for time to failure with corresponding cumulative distribution function given by,
\[ F(t) = 1 - e^{-at^b}, \quad a > 0, b > 0 \]
Which ONE of the following represents \( r(t) \)?
Show Hint
Differentiate \( F(t) \) to get \( f(t) \), then divide by \( 1-F(t) \); the exponential terms cancel.
Updated On: Aug 5, 2026
- \( a \times b \times t^{b-1} \times e^{-at^b} \)
- \( a \times b \times t^{b-1} \)
- \( 1 - (a \times b \times t^{b-1}) \)
- \( 1 - (a \times b \times t^{b-1} \times e^{-at^b}) \)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Recognize the distribution:
The CDF $F(t) = 1 - e^{-at^b}$ is the standard form of the Weibull distribution, with $a$ acting as a scale-related constant and $b$ as the shape parameter.
This distribution is used heavily in reliability analysis because its hazard rate can increase, decrease, or stay constant depending on $b$.
Step 2: Apply the hazard rate definition:
By definition the hazard function is the density divided by the survival probability.
\[ r(t) = \frac{f(t)}{1 - F(t)} \]
Here $f(t) = F'(t)$, so before substituting we need the derivative of $F(t)$ with respect to $t$.
Step 3: Work out $f(t)$ and $1 - F(t)$ separately:
Differentiating $F(t) = 1 - e^{-at^b}$ with respect to $t$:
\[ f(t) = a b t^{b-1} e^{-at^b} \]
And directly from the given CDF:
\[ 1 - F(t) = e^{-at^b} \]
Dividing the two:
\[ r(t) = \frac{a b t^{b-1} e^{-at^b}}{e^{-at^b}} = a b t^{b-1} \]
Step 4: Confirm with the Weibull property:
A known fact about the Weibull distribution is that its hazard function always reduces to a pure power of $t$, scaled by shape and scale constants, never carrying the exponential term.
This matches what we derived, and rules out options (A), (C), and (D), each of which either keeps the exponential or introduces an unjustified subtraction from 1.
Final Answer:
The hazard rate simplifies neatly once the exponential terms cancel. \[ \boxed{r(t) = a b t^{b-1}} \]
The CDF $F(t) = 1 - e^{-at^b}$ is the standard form of the Weibull distribution, with $a$ acting as a scale-related constant and $b$ as the shape parameter.
This distribution is used heavily in reliability analysis because its hazard rate can increase, decrease, or stay constant depending on $b$.
Step 2: Apply the hazard rate definition:
By definition the hazard function is the density divided by the survival probability.
\[ r(t) = \frac{f(t)}{1 - F(t)} \]
Here $f(t) = F'(t)$, so before substituting we need the derivative of $F(t)$ with respect to $t$.
Step 3: Work out $f(t)$ and $1 - F(t)$ separately:
Differentiating $F(t) = 1 - e^{-at^b}$ with respect to $t$:
\[ f(t) = a b t^{b-1} e^{-at^b} \]
And directly from the given CDF:
\[ 1 - F(t) = e^{-at^b} \]
Dividing the two:
\[ r(t) = \frac{a b t^{b-1} e^{-at^b}}{e^{-at^b}} = a b t^{b-1} \]
Step 4: Confirm with the Weibull property:
A known fact about the Weibull distribution is that its hazard function always reduces to a pure power of $t$, scaled by shape and scale constants, never carrying the exponential term.
This matches what we derived, and rules out options (A), (C), and (D), each of which either keeps the exponential or introduces an unjustified subtraction from 1.
Final Answer:
The hazard rate simplifies neatly once the exponential terms cancel. \[ \boxed{r(t) = a b t^{b-1}} \]
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