Respuesta :
Answer:
a) [tex]\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}[/tex]
b) [tex]P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736[/tex]
Step-by-step explanation:
Part a
We assume that the parachutist lands at random point in the interval (x=A,y=B) we have a continuous random variable X. And the distribution of X would be uniform [tex]Y\sim Unif(A,B)[/tex]. And the density function would be given by:
[tex]f(x) =\frac{1}{B-A} , A <x<B[/tex]
And 0 for other case.
The operation fails, if parachutist's distance to A is more than five times as much as her distance to B.
So the point P in the interval (A,B) at which the distance to A is exactly 5 times the distance to B is given by:
[tex]P= A + \frac{5}{6} (B-A)= \frac{6A +5B -5A}{6}=\frac{A+5B}{6}[/tex]
And we can find the probability desired like this:
[tex]P(d(P,A) \geq 5 d(P,B))= P(\frac{A+5B}{6} < X< B)[/tex]
And from the cumulative distribution function of X ficen by [tex]F(X)\frac{X-A}{B-A}[/tex] we got:
[tex]\frac{B-\frac{A+5B}{6}}{B-A}= \frac{6B -A-5B}{6(B-A)}=\frac{B-A}{6(B-A)}=\frac{1}{6}[/tex]
Part b
For this case we assume that [tex]X\sim Gamma (2,1)[/tex]
On this case we assume that [tex]\alpha=2, \beta= 1[/tex]
The density function for the Gamma distribution is given by:
[tex]P(X)= \frac{\beta^{\alpha} x^{\alpha-1} e^{-\beta x}}{\gamma(\alpha)}[/tex]
And on this case we can find the probability using the complement rule like this:
[tex]P(X>1) = 1-P(X\leq 1)=0.736[/tex]
We can solve this problem with the following excel code:
"=1-GAMMA.DIST(1;2;1;TRUE)"
And if we do it by hand we need to do this:
[tex]P(X>1) = 1-P(X\leq 1)=1-\int_{0}^1 \frac{1^{2} x^{2-1} e^{- x}}{\gamma(2)}=0.736[/tex]
