Question

The number of days ahead travelers purchase their airline tickets can be modeled by an exponential distribution with the average amount of time equal to 15 days. Find the probability that a traveler will purchase a ticket fewer than ten days in advance. How many days do half of all travelers wait

Answer 1

Answer:

The right answer is:

(a) 0.4866

(b) 10.3972

Step-by-step explanation:

Let X be the number of days,

X is $exp(\lambda)$,

⇒ $\lambda =\frac{1}{15}$

now,

⇒ $F_x(x) = P(X \leq x)$

              $=1-e^{-\lambda x}$

              $=1-e^{-\frac{x}{15} }$

or,

$x>0$

(a)

⇒ $P(X<10)=F_x(10)$

                      $=1-e^{\frac{10}{15} }$

                      $=1-e^{\frac{2}{3} }$

                      $=0.4866$

(b)

If the medium be m, then

$P(X=m) = \frac{1}{2}$

                 $=2^{-1}$

⇒$1-e^{-\frac{m}{15} } = \frac{1}{2}$

        $e^{-\frac{m}{15} } = 2^{-1}$

             $m= 15\ ln2$

                 $=10.3972$