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  1. #1
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    probability question




    Hello,
    I am having trouble with my stat homework. Maybe someone can give me some guidance? I would really appreciate it!

    A woman is fishing for bass on Pinelands pond. Let x=number of 'keepers' that she catches in an hour. The probability distribution for x is given in the table below

    x= 0 1 2 3 4 5
    p(x)= .1 .2 .3 .2 .1 .1


    Questions:
    a) What is the probability that she catches at most 4 keepers?
    b) What is the probability that x is greater to and equal to 3?

    c) Find Ux, the mean number of keepers caught
    d) Find the standard deviation for X, the number of keepers caught

  2. #2
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    a) the probability of at most 4 is equal to the probability of three or fewer. simply add up the probabilities involved.

    b) use the above and adjust for three or greater

    some examples:

    c) for a probability distribution x= 2 4 6 8 10, P(x)= .1 .3 .1 .2 .3 Ux + 2(.1)+4(.3)+6(.1)+8(.2)+10(.3) = 6.6

    d) sigma = sqrt( sum(x-mu)^2*P(x)) so, to begin the pattern for my example in c) = sqrt((2-6.6)^2(.1) + (4-6.6)^2(.3)...)=2.83

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    c)
    this is what i did, tell me if i am correct?
    0(.1) + 1(.2) + 2(.3) + 3(.2) + 4(.1) + 5(.1)
    mean = 2.3

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    a)
    .1 + .2 + .3 + .2 + .1 = .90
    is this correct also?

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    b)
    .2 + .1 + .1 = .40
    is this correct also?

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    d)
    is the standard deviation 7.3?
    please someone check my answers. Thank you very much!

  7. #7
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    Check your d).
    abc looks fine.
    In the long run, we're all dead.

  8. #8
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    Yeah im confused on d)
    I did this:

    0^2(.1) + 1^2(.2) + 2^2(.3) + 3^2(.2) + 4^2(.1) + 5^2(.1)

    what did i do wrong??

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    follow the formula
    sigma = sqrt( sum {( x-mu)^2*P(x) } ) as jerry mentioned

    mu is the mean
    In the long run, we're all dead.

  10. #10
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    alright i think i got it
    0^2(.1) + 1^2(.2) + 2^2(.3) + 3^2(.2) + 4^2(.1) + 5^2(.1)

    0+.2+1.2+1.8+1.6+2.5

    =7.3 - mean squared 5.29 = 2.01

    did i do it?!

  11. #11
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    correct. But it is not the sd.
    you have calculated the variance.
    In the long run, we're all dead.

  12. #12
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    i am terrible at statistics...ugh
    back to the drawing board.

  13. #13
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    okay i calculated it again but came out to 2.01!!! which is the variance.
    if i squared the variance (2.01)^2 = 4.0401
    would that be the standard deviation??
    thanks

  14. #14
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    Read the text book thoroughly.

    sd = sqrt(var)
    In the long run, we're all dead.

  15. #15
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    so you are saying...
    take the square root of 2.01
    which is 1.4177 ...is the standard deviation??

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