Question
An instrument registers one signal during the ten seconds after it is switched on. Its signal time \(T\) is continuous and has a density that is three times as high during the first two seconds as it is during the following eight seconds. More precisely,
\(f(t)=3c\) for \(0\le t<2\); \(f(t)=c\) for \(2\le t\le10\); \(f(t)=0\) otherwise,
where \(c\) is an unknown positive constant with units of inverse seconds.
Find \(c\), then calculate \(\Pr(1\le T\le5)\). Explain why the higher density in the first band does not mean that band is more likely overall.
Get hint
Hint 1
Probability is area under the density. Multiply each constant height by the width of its band, then require the total area to equal 1.
Check your answer
Solution
A probability density must integrate to 1. The first band has width 2 seconds and height \(3c\); the second has width 8 seconds and height \(c\). Hence
\(\int f(t)\,dt=(2)(3c)+(8)(c)=14c=1\),
so \(c=\frac1{14}\,\mathrm{s}^{-1}\).
The event \(1\le T\le5\) straddles the density change at \(t=2\), so compute both pieces:
\[\begin{aligned}P(1\le T\le5)&=\int_1^2 3c\,dt+\int_2^5 c\,dt\\&=3c(2-1)+c(5-2)\\&=6c=\mathbf{6/14=3/7}.\end{aligned}\]
For the entire first band, \(\Pr(0\le T<2)=2\cdot3c=\frac6{14}=\frac37\); for the second, \(\Pr(2\le T\le10)=8\cdot c=\frac8{14}=\frac47\). The first band's density is higher but it lasts only two seconds instead of eight. Probability is height times width, not density height alone. The probability of the single boundary point \(T=2\) is zero, so the choice of open or closed endpoint there does not affect the answer.
Assumptions
T is continuous with the exact piecewise constant density stated, zero outside [0,10].
Band endpoints have probability zero.
All time inputs are measured in seconds and c is the same scaling constant in both bands.
Final Answer
\(c=\frac1{14}\) per second; \(\Pr(1\le T\le5)=\frac37\). The first band has probability 3/7, while the longer second band has probability 4/7.