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stuck with this question (1 Viewer)
- Thread starter NexusRich
- Start date
P(X<50) = 0
P(X<= 60) = P(50 <= X <= 60)
P(X <= 75) = P(50 <= X <= 75)
P(X >= 65) = 1 - P(50 <= X < 65)
P(X >= 99) = 1 - P(50 <= X < 99)
For a <= 50, P(X <= a) = 0
For a > 50:
P(X <= a) = P(50 <= X < a) and:
 = \int _{50} ^a f(x) dx = \int _{50} ^a (15000x^{-3} - 750000x^{-4})dx \\ \\ = \left[ \frac{15000x^{-2}}{-2} - \frac {750000x^{-3}}{-3} \right]_{50} ^a = \left [ \frac{250000}{x^3} - \frac {7500}{x^2}\right]^a _{50} \\ \\ \therefore P(X \leq 60) = \left[( \cdots) \right ]^{60} _{50} = 0.074074 . . . ??? \\ \\ P(X \leq 75) = \left [ ( \cdots) \right ]^{75} _{50} = ???)
You should be able to continue from there.
I may have made some mistakes in computation above. No time to complete, sorry.
P(X<= 60) = P(50 <= X <= 60)
P(X <= 75) = P(50 <= X <= 75)
P(X >= 65) = 1 - P(50 <= X < 65)
P(X >= 99) = 1 - P(50 <= X < 99)
For a <= 50, P(X <= a) = 0
For a > 50:
P(X <= a) = P(50 <= X < a) and:
You should be able to continue from there.
I may have made some mistakes in computation above. No time to complete, sorry.
Last edited:
5uckerberg
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- 2018
Note for probability you should always have a positive value which can be shown because the upper boundary iscan someone pls help with part b and c, I tried it a few times and kept getting the wrong answer
If they are asking for the other side which isthen this is our attack pattern
dx)
The rest which is integration is taking candy from a baby.