The pdf for the lifetime X, in years, of a Superstuff disk drive is given as follows:
$f(x) = \begin{cases} 2/x^2 & \text{for } x\geq2\text{ } \\ 0 & \text{elsewhere} \end{cases}$.
(i) Determine the cdf, $F(x)$ for all $x$ values.
My attempt:
$F(x_0)$ $=$ $ \int_{-\infty}^{2} 0dx +2 \int_{2}^{x_0} x^{-2} dx$
$=$ $1-(2/x_0)$
Answer:
$F(x)$ $=$ $0$ for $x<2$
$F(x)$ $=$ $(-2/x) + 1$ for $x\geq 2$
(ii) Find the probability $P(3\leq X \leq 4)$
$P(3<x<4)=$ $\int_{3}^{4}f(x) dx=2\int_{3}^{4}x^{-2}dx$
Answer:
$1/6$
(iii) Determine $x_1$ such that $P(X<x_1)=2/3$ (Here $x_1$ is the $66.7$th percentile)
I am not sure about this one. Can someone please show me how to do this with the working?
(iv) Out of a set of $10$ randomly selected Superstuff disk drives, what is the probability that at least $4$ will last longer than $5$ years?
$P(X\geq 5$ $=$ $\int_{5}^{\infty} 2/x^{2}dx$
$=$ $2/5$
$P(X\geq 4)= 1-P(X<4)=$
$=$ $1-$${10\choose 3}(2/5)^3(3/5)^7 + {10\choose 2}(2/5)^2(3/5)^8 + {10 \choose 1}(2/5)^1(3/5)^9 + {10 \choose 0}(2/5)^0(3/5)^{10}\;$
Answer:
$=$ $.6178$
Is this correct? If so, is there another way of doing this problem? I am not sure about (iii). Can someone please help me with this and the working as well?