Respuesta :
Answer:
a) [tex] f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11[/tex]
b) [tex] P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286[/tex]
c) [tex]P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857[/tex]
d) [tex] P(X >5 | X \leq 7)[/tex]
And we can find this probability with this formula from the Bayes theorem:
[tex] P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5 <X \leq 7)}{P(X<7)}=\frac{0.2857}{0.4286}= 0.6666 [/tex]
Step-by-step explanation:
For this case we assume that the random variable X follows this distribution:
[tex] X \sim Unif (a=4, b =11)[/tex]
Part a
The probability density function is given by the following expression:
[tex] f(x) = \frac{1}{b-a} , a \leq x \leq b[/tex]
[tex] f(x)= \frac{1}{11-4}= \frac{1}{7}, 4 \leq x \leq 11[/tex]
Part b
We want this probability:
[tex] P(X \leq 7)[/tex]
And we can use the cumulative distribution function given by:
[tex] F(x) = \frac{x-a}{b-a}= \frac{x-4}{11-4} [/tex]
And replacing we got:
[tex] P(X\leq 7) = F(7) = \frac{7-4}{11-4}= 0.4286[/tex]
Part c
We want this probability:
[tex]P(5 < X \leq 7)[/tex]
And we can use the CDF again and we have:
[tex]P(5 < X \leq 7)= F(7) -F(5) = \frac{7-4}{7} -\frac{5-4}{7}= 0.2857[/tex]
Part d
We want this conditional probabilty:
[tex] P(X >5 | X \leq 7)[/tex]
And we can find this probability with this formula from the Bayes theorem:
[tex] P(X >5 | X \leq 7)= \frac{P(X>5 \cap X \leq 7)}{P(X \leq 7)}= \frac{P(5 <X \leq 7)}{P(X<7)}=\frac{0.2857}{0.4286}= 0.6666 [/tex]
