Respuesta :
Answer:
a)[tex] P(14\leq X \leq 26) = \frac{12}{21}[/tex]
b) [tex]P(18\leq X \leq 22) = \frac{4}{21}[/tex]
c) [tex]P(X\leq 19) = P(14\leq X \leq 19) = \frac{5}{21}[/tex]
d) [tex]P(\{X=19\}\cup\{X=56\}) = 0[/tex]
Step-by-step explanation:
Let X be the speed of a car. So, we know that X is distributed uniformly over the interval [14,35]. Recall that if X is distributed uniformly over the set [a,b] then the density function of X is
[tex] f(x) = \frac{b-a}[/tex if x is in the set [a,b] and 0 otherwise.
So in our case, the density function is [tex] f(x) = \frac{1}{21}[/tex]. Using this function we have that
[tex]\text{P}(a \leq X\leq b) = \int_{a}^{b} f(x) dx[/tex]. In this case
[tex]P(a\leq X \leq b = \int_{a}^{b} \frac{1}{21}dx = \frac{b-a}{21}[/tex] when a,b are in [14,35].
a).We are asked for [tex]P(X\leq 26) = P(14\leq X \leq 26) = \frac{26-14}{21}=\frac{12}{21}[/tex]
b) [tex]P(18\leq X \leq 22) = \frac{4}{21}[/tex]
c) Note that since 35 < 56 it is always true that the speed is under 56. So, the probability that the speed of the car is less than 19 given that the speed is below 56 is the same as having the probability that the speed is less than 19. That is
[tex]P(X\leq 19) = P(14\leq X \leq 19) = \frac{5}{21}[/tex]
d) It is impossible that the speed is 19 and 56 at the same time, so the probability that the speed is 19 and 56 is 0. Since the speed is at most 35, it is impossible that the speed is 56. So the probability of having that the speed is 56 is 0. Recall from the definition that
[tex] P(X=19) = P(19\leq X \leq 19) = \frac{19-19}{21} = 0[/tex]
So, using the following formula for events A,B
[tex]P(A\cup B) = P(A)+P(B) - P(A\cap B) [/tex]
then,
[tex]P(\{X=19\}\cup\{X=56\}) = 0[/tex]
