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Suppose that the life distribution of an item has hazard rate function $\lambda(t)=1.5t^2$, $t>0$. What is the probability that:

  1. The item doesn't survive to age $2$?

  2. The item's lifetime is between $0.5$ and $4$?

  3. A $1.5$ year-old item will survive to age $2$?

For (1) I got the correct answer! $$1 - e^{-.5(2^3)}$$

For (2) Here is what I keep getting: $e^{-.5(64)}- e^{-.5(.125)}$ The $64$ comes from $4^3$ and the $.125$ comes from $.5^3$.

(3) This is the answer I got, but it is also wrong: $\frac{e^{-.5((1.5)^3)}}{e^{-.5}}$

Can someone help correct me/show me what to do?

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  • $\begingroup$ What does 1.5t2 mean? $\endgroup$ – barak manos Nov 22 '14 at 20:48
  • $\begingroup$ Sorry 1.5t^2 is what i meant $\endgroup$ – Kris Nov 22 '14 at 20:53
  • $\begingroup$ On another point, I strongly suggest you re-type your question using mathjax, as it would make it easier for others to read your question and help you accordingly. $\endgroup$ – alonso s Nov 22 '14 at 21:01
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    $\begingroup$ @alonsos I just put in an edit with proper formatting, just needs to be approved. $\endgroup$ – Rivasa Nov 22 '14 at 21:02
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If the survival function is as suggested by your first answer $$S(t)=P(T \gt t) =\exp\left(-\frac{t^3}{2}\right)$$ then your second answer should be $S(0.5)-S(4)$ but you seem to have $S(4)-S(0.5)$,

and your third answer should be $\dfrac{S(2)}{S(1.5)}$ but you seem to have $\dfrac{S(1.5)}{S(1)}$.

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  • $\begingroup$ Ok I see what I did wrong for c but for part b, why is it switched? $\endgroup$ – Kris Nov 22 '14 at 21:15
  • $\begingroup$ You are more likely to survive at least $0.5$ years than at least $4$ years. If you had the cumulative distribution function $F(x)=\Pr(T \le t)=1-\exp\left(-\frac{t^3}{2}\right)$ then you would be correct to calculate $F(4)-F(0.5)$. $\endgroup$ – Henry Nov 22 '14 at 21:32

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