Exponential probability density functions

Part a)

Using technology: $\displaystyle \int_0^\infty0.05e^{-0.05t}\,dt=1$

and $f(t)\ge0$ for all $t$

Part b)

$\displaystyle P(T<10)=\int_0^{10}0.05e^{-0.05t}\,dt$

Using technology: $\approx0.39$

Part c)

$\displaystyle \int_5^m0.05e^{-0.05t}\,dt=0.6$

Using solve facility of technology: $m\approx34.43$ hours

Part a) needs the integral from $0$ to $\infty$. An integral over a finite interval doesn't get the mark.

Part a)

$\displaystyle \lambda=\frac18=0.125$, $f(t)=0.125e^{-0.125t}$, $t\ge0$

Part b)

$\displaystyle P(4<T<10)=\int_4^{10}0.125e^{-0.125t}\,dt$

Using technology: $\approx0.32$

Part c)

$\displaystyle \int_0^M0.125e^{-0.125t}\,dt=0.5$

Using solve facility of technology: $M\approx5.55$ minutes ($8\ln(2)$)

Part a)

$\displaystyle P(T>6)=\int_6^\infty\lambda e^{-\lambda t}\,dt=e^{-6\lambda}=0.25$

$\displaystyle \lambda=\frac{\ln(4)}{6}\approx0.2310$

Part b)

$\displaystyle \sigma=\frac1\lambda\approx4.33$ years

Part c)

$e^{-0.2310m}=0.1$

$\displaystyle m=\frac{\ln(10)}{0.2310}\approx9.97$ years

Part a)

$\displaystyle P(T\le b)=\int_0^b\lambda e^{-\lambda t}\,dt=\Big[-e^{-\lambda t}\Big]_0^b$

$=-e^{-\lambda b}-(-1)=1-e^{-\lambda b}$

Part b)

Median $M$: $1-e^{-\lambda M}=\tfrac12\Rightarrow e^{-\lambda M}=\tfrac12$

$\displaystyle -\lambda M=\ln\left(\frac12\right)=-\ln(2)\Rightarrow M=\frac{\ln(2)}{\lambda}$

Part c)

$\lambda=\tfrac15$, $P(T>10)=1-\left(1-e^{-\frac{10}{5}}\right)=e^{-2}$

Part a)

$\displaystyle P(T>3\ln(4))=\int_{3\ln(4)}^\infty\frac13e^{-\frac{t}{3}}\,dt=\Big[-e^{-\frac{t}{3}}\Big]_{3\ln(4)}^\infty$

$=0+e^{-\ln(4)}=\dfrac14$

Part b)

$1-e^{-\frac{q}{3}}=0.9\Rightarrow e^{-\frac{q}{3}}=0.1$

$-\tfrac{q}{3}=\ln(0.1)=-\ln(10)\Rightarrow q=3\ln(10)$

Part c)

Mean $=\tfrac1\lambda$: $\displaystyle P\left(T>\frac1\lambda\right)=e^{-\lambda\times\frac1\lambda}=e^{-1}$, which does not depend on $\lambda$.

$\ln(1000)$ is the same as $3\ln(10)$ and is also accepted.

Part a)

$\lambda=\tfrac14$, $P(T>4)=e^{-1}\approx0.368$

Approximately 36.8% of watches last longer than the mean.

Part b)

Warranty $w$ years: $P(T<w)=1-e^{-0.25w}\le0.05$

$\displaystyle w\le\frac{-\ln(0.95)}{0.25}\approx0.205$ years $\approx2.46$ months

Rounding up to 3 months would replace more than 5%, so the longest warranty is 2 months.

Part c)

The distribution is right-skewed (a long tail of large lifetimes), which pulls the mean above the median ($\approx2.77$ years compared with 4 years).

Rounding 2.46 months to the nearest month gives 3 months, which would replace more than 5% of watches and loses the third mark. Round down here.

Part a)

Let $u=t\Rightarrow\dfrac{du}{dt}=1$, $\dfrac{dv}{dt}=\lambda e^{-\lambda t}\Rightarrow v=-e^{-\lambda t}$

$\displaystyle \int_0^b\lambda te^{-\lambda t}\,dt=\Big[-te^{-\lambda t}\Big]_0^b+\int_0^be^{-\lambda t}\,dt=-be^{-\lambda b}+\Big[-\frac1\lambda e^{-\lambda t}\Big]_0^b$

$\displaystyle =-be^{-\lambda b}-\frac1\lambda e^{-\lambda b}+\frac1\lambda=\frac1\lambda-e^{-\lambda b}\left(b+\frac1\lambda\right)$

Part b)

$\displaystyle E(T)=\int_0^\infty\lambda te^{-\lambda t}\,dt$. As $b\to\infty$, $e^{-\lambda b}\to0$ and $be^{-\lambda b}\to0$, so $E(T)=\dfrac1\lambda$.

Part c)

Mean $=2$: $P(T<2)=1-e^{-0.5\times2}=1-e^{-1}$

Part a)

$P(T_B<2)=1-e^{-2\lambda}=0.3\Rightarrow e^{-2\lambda}=0.7$

$\displaystyle \lambda=\frac{-\ln(0.7)}{2}\approx0.1783$

Part b)

Brand A: $\lambda=\tfrac16$, $P(T_A>5)=e^{-\frac56}\approx0.43$

Brand B: $P(T_B>5)=e^{-0.1783\times5}\approx0.41$

Brand A has the greater probability of lasting more than 5 years.

Part c)

$1-e^{-0.1783t}=0.9\Rightarrow t=\dfrac{\ln(10)}{0.1783}\approx12.91$ years

$P(T>5)=e^{-5\lambda}=0.4$

$\displaystyle \lambda=\frac{-\ln(0.4)}{5}\approx0.1833$

$\displaystyle P(T<3)=\int_0^3\lambda e^{-\lambda t}\,dt$

Using technology: $P(T<3)\approx0.42$

$0.42<0.5$, so fewer than half of the times between calls are shorter than 3 minutes. (Equivalently, the median $\tfrac{\ln(2)}{\lambda}\approx3.78$ minutes is greater than 3 minutes.)

The manager's claim is not reasonable.

Read the records carefully: 40% are longer than 5 minutes. Writing $P(T<5)=0.4$ instead gives $\lambda\approx0.1022$ and loses the first two marks, but you can still get the rest.

$\displaystyle \int_0^\infty ke^{-\frac{t}{2}}\,dt=1\Rightarrow k\Big[-2e^{-\frac{t}{2}}\Big]_0^\infty=1$

$2k=1\Rightarrow k=\tfrac12$

$\displaystyle P(a<T<2a)=\Big[-e^{-\frac{t}{2}}\Big]_a^{2a}=e^{-\frac{a}{2}}-e^{-a}=\frac14$

Let $u=e^{-\frac{a}{2}}$, so $e^{-a}=u^2$:

$u-u^2=\tfrac14\Rightarrow u^2-u+\tfrac14=0\Rightarrow\left(u-\tfrac12\right)^2=0$

$u=\tfrac12\Rightarrow e^{-\frac{a}{2}}=\tfrac12\Rightarrow a=2\ln(2)$

$k=\tfrac12$, $a=2\ln(2)$

You can also say straight away that $f(t)$ is exponential with $\lambda=\frac12$, so $k=\frac12$, as long as you give that reason. $\ln(4)$ is the same as $2\ln(2)$.