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mariomintchev

  • 2 years ago

Help with #6 on the worksheet....

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

  3. mariomintchev
    • 2 years 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
    • 2 years ago
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    @robtobey @phi

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

  6. mariomintchev
    • 2 years ago
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    hi @Jemurray3 ! :D

  7. mariomintchev
    • 2 years ago
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    **** c) Pr(X<160)

  8. Jemurray3
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    for D would it be 2 times the numerator??

  11. Jemurray3
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    so 1/6 times 1/6 ??

  14. Jemurray3
    • 2 years ago
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    Yes

  15. mariomintchev
    • 2 years ago
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    sweeeeet

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

  17. Jemurray3
    • 2 years ago
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    yes

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

  19. mariomintchev
    • 2 years ago
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    @Jemurray3 , you still alive? haha

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

  21. mariomintchev
    • 2 years ago
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    ok. im just making sure.

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