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mariomintchev

  • one year ago

Help with #6 on the worksheet....

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  1. mariomintchev
    • one year ago
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  2. mariomintchev
    • one year ago
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    this is what i have: |dw:1360104752412:dw|

  3. mariomintchev
    • one year ago
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    b) Expected Value (Height) = ? c) Pr(X>160) d) e) Pr(165<x<185) f) Pr(X>185) Help????

  4. mariomintchev
    • one year ago
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    @robtobey @phi

  5. mariomintchev
    • one year ago
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    ok b) is 1/(b-a) so 1/(210-150)

  6. mariomintchev
    • one year ago
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    hi @Jemurray3 ! :D

  7. mariomintchev
    • one year ago
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    **** c) Pr(X<160)

  8. Jemurray3
    • one year ago
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    What you said above, 1/(b-a), is the probability density function: P(x) = 1/(b-a) for all x between a and b, and zero elsewhere. Because it's constant, you can multiply it by the length of any interval to find the probability that someone's height falls in said interval. So, (c) is (160-150) / (210-150) = 10/60 = 1/6.

  9. mariomintchev
    • one year ago
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    o ok i see... please continue.... my notes have what you just said but are not very clear....

  10. mariomintchev
    • one year ago
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    for D would it be 2 times the numerator??

  11. Jemurray3
    • one year ago
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    The expectation value is fairly clearly the average height. The rest are just multiplying your probability density function by intervals of different lengths.

  12. Jemurray3
    • one year ago
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    No, it is not. If A and B are uncorrelated random events, then P(A and B) = P(A) * P(B)

  13. mariomintchev
    • one year ago
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    so 1/6 times 1/6 ??

  14. Jemurray3
    • one year ago
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    Yes

  15. mariomintchev
    • one year ago
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    sweeeeet

  16. mariomintchev
    • one year ago
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    for part E would i do: (185-165)/(210-150) ??

  17. Jemurray3
    • one year ago
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    yes

  18. mariomintchev
    • one year ago
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    and F would be (210-200)/(210-150) ?? lol

  19. mariomintchev
    • one year ago
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    @Jemurray3 , you still alive? haha

  20. Jemurray3
    • one year ago
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    You seem to understand it well enough without me at this point

  21. mariomintchev
    • one year ago
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    ok. im just making sure.

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