Problem: The mileage $C$ in thousands of kilometers which car owners get with a certain kind of tyres is a random variable having probability density function
\[
f(x) =
\begin{cases}
\frac{1}{20}e^{-\frac{x}{20}}, &\text{ if } x > 0;\\
0, &\text{ if } x \leq 0.
\end{cases}
\]
Find the probabilities that one of these tyres will last
-
at most $10,000$ kilometers;
-
anywhere from $16,000$ to $24,000$ kilometers;
-
at least $30,000$ kilometers.
Solution: The given probability distribution function is
\[
f(x) =
\begin{cases}
\frac{1}{20}e^{-\frac{x}{20}}, &\text{ if } x > 0;\\
0, &\text{ if } x \leq 0.
\end{cases}
\]
Let $X$ denote the millage in thousand kilometers with a certain kind of tyre.
-
We need to find the probability that one of the tyres will last at most $10,000$ kilometers. It will be
\begin{align*}
P( X \leq 10) & = \int _{-\infty }^{10} f(x) \mathrm{d} x\\
& = \int _0^{10} \frac{1}{20} e^{-\frac{x}{20}} \mathrm{d} x \\[1ex]
& = \frac{1}{20} \left[ \dfrac{e^{-\frac{x}{20}}}{-1 / 20} \right]_0^{10} \\[1ex]
& = \left[ -e^{-\frac{10}{20}} + e^{\frac{0}{20}} \right] \\[1ex]
& = 1 - e^{-\frac{1}{2}} = 0.3935.
\end{align*}
-
The probability that one of the tyres will last between $16,000$ to $24,000$ kilometers will be
\begin{align*}
P(16 \le X \le 24) & = \int_{16}^{24} f(x) \mathrm{d} x \\
& = \frac{1}{20} \int_{16}^{24} e^{-\frac{x}{20}} \mathrm{d} x \\[1ex]
& = \frac{1}{20}\left[ \dfrac{e^{-\frac{x}{20}}}{-1 / 20} \right]_{16}^{24} \\[1ex]
& = \left[ - e^{-\frac{24}{20}} + e^{-\frac{16}{20}} \right] \\[1ex]
& = e^{-\frac{4}{5}} - e^{-\frac{6}{5}} \\
& = 0.1481.
\end{align*}
-
Finally, the probability that one of the tyres will last at least $30,000$ kilometers will be
\begin{align*}
P(X \geq 30) & = \int_{30}^{\infty} f(x)\mathrm{d} x\\
& = \frac{1}{20} \int_{30}^{\infty } e^{-\frac{x}{20}} \mathrm{d} x\\[1ex]
& = \frac{1}{20} \left[ \dfrac{e^{-\frac{x}{20}}}{-1 /20} \right]_{30}^{\infty } \\[1ex]
& = e^{-\frac{30}{20}} = 0.2231.
\end{align*}