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- Thread starter NexusRich
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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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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
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The rest which is integration is taking candy from a baby.