Thread: probability density

the trend wear in thousands of k, thatt car owners get with a kind of tire is a RV whose probability density is

f(x) = 1.30 exp(-x/30) for x>- and 0 for x less than or equal to 0.

find the probabilities that one of these tires will last
(a)at most 18,000 km.
(b) between 27, 000 to 36,000 km
(c) at least 48, 000km

Should it have been;

Originally Posted by alexandrabel90

f(x) = 1/30 exp(-x/30) for x>- and 0 for x less than or equal to 0.

find the probabilities that one of these tires will last
(a)at most 18,000 km.
(b) between 27, 000 to 36,000 km
(c) at least 48, 000km

?

if so,
(a) $\displaystyle \Pr(0<X<18000)=\int_{0}^{18} \frac{1}{30}e^{-\frac{x}{30}}=\left[ -E^{(-x/30)}\right]^{18}_{0}=1-\frac{1}{e^{3/5}}=0.45118$
(b) $\displaystyle \Pr(27<X<36)=\int_{36}^{27} \frac{1}{30}e^{-\frac{x}{30}}$
(c)$\displaystyle \Pr(X>48)=1-\Pr(X<48)=1-\int_{0}^{48} \frac{1}{30}e^{-\frac{x}{30}}$