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Math Help - Assignment 3 help..!!! so hard......!!!!

  1. #1
    Newbie sonymd23's Avatar
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    Assignment 3 help..!!! so hard......!!!!

    i just finish the home work........can any one check my work correct me if i didnt get it right, i only done the questions that i can solve.....and can any one do question #4,9,10,12 (b,c) ,13,14,15,16 <- those are the one i dont know how to solve .........









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  2. #2
    MHF Contributor Quick's Avatar
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    where is the second page of assignment three?
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  3. #3
    Newbie sonymd23's Avatar
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    Quote Originally Posted by Quick
    where is the second page of assignment three?

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  4. #4
    MHF Contributor Quick's Avatar
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    That's the first page of assignment 2...
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  5. #5
    Newbie sonymd23's Avatar
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    Quote Originally Posted by Quick
    That's the first page of assignment 2...

    no is not.....look at the top left corner....it say assignment 3 page 2......however can u do ur best....and ASAP

    by the one.....how come u keep saying assig 2 ? i dont get it....?
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  6. #6
    Grand Panjandrum
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    Question 4.

    (a) If for two events A and B

    P(B|A)=P(B),

    then as:

    P(B)=P(B|A)+P(B| \mathrm{not }A)

    we have:

    P(B| \mathrm{not }A)=0.

    (b) A dice is rolled twice. The first roll is a 2, what is the probability
    that the sum of the two rolls is greater than 6?

    P(a+b>6|a=2)=P(b=5)+P(b=6)

    that is the probability of the sum being greater than 6 given the first roll
    is a 2, is the sum of the probabilities that the second roll is a 5 and that
    the second roll is 6. So:

    P(a+b>6|a=2)=1/6+1/6=2/6=1/3

    RonL
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  7. #7
    Eater of Worlds
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    #16:

    \hat{p}=.62\;\ and \;\  \hat{q}=.38

    From the table, the z-score for a 92% CI is 1.75.

    E=1.75\sqrt{\frac{(.62)(.38)}{100}}\approx{.085}

    .62+.085=.705

    .62-.085=.535

    .535<p<.705

    You can say with 92% confidence that the proportion of people who buy SNO flakes is between 53.5% and 70.5%.
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    Sonymd23 do not beg. If you question is not answered do not start posting totally useless posts begging for answers. Wait.
    -=USER WARNED=-
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  9. #9
    Newbie sonymd23's Avatar
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    Quote Originally Posted by galactus
    #16:

    \hat{p}=.62\;\ and \;\  \hat{q}=.38

    From the table, the z-score for a 92% CI is 1.75.

    E=1.75\sqrt{\frac{(.62)(.38)}{100}}\approx{.085}

    .62+.085=.705

    .62-.085=.535

    .535<p<.705

    You can say with 92% confidence that the proportion of people who buy SNO flakes is between 53.5% and 70.5%.
    how did u get the 1.75?
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  10. #10
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    how did u get the 1.75?
    Virus.BagelJQ.Dll
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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by sonymd23
    how did u get the 1.75?
    Since you are looking for a 92% CI the upper limit must cut off 4% of
    the area under the normal density above the limit (the lower limit will cut off
    the other 4% below, to leave the 92% you require). So the area under the
    normal density curve below the upper limit is 96% which is what you look up
    in a standard normal table to get 1.75.

    RonL
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  12. #12
    Eater of Worlds
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    Quote Originally Posted by sonymd23
    how did u get the 1.75?
    From the standard normal distribution table, aka the z-table.

    I calculated 92% CI by looking up .96 in the table and seeing what z-score sorresponded.


    You should see your teacher.
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  13. #13
    Newbie sonymd23's Avatar
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    ^ thnx every one......however i dont get number #13........

    explain, using a sketch graph, why the continuity correction is needed for discrete data. use an example to illustrate your explanation.
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  14. #14
    Eater of Worlds
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    When you use a continuous normal distribution to approximate a binomial probability, you need to move 0.5 units to the left and right of the midpoint. This is so you can include all possible x-values in the interval.

    This is called a continuity correction

    For instance, let's convert the following binomial to normal distribution.

    The probability of getting between 200 and 250 successes, inclusive.

    The discrete midpoint values are 200, 201, 202,....., 250. The interval for the

    normal distribution is 199.5<x<250.5
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  15. #15
    Newbie sonymd23's Avatar
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    ^ thnx........but do u have a graph ?

    beucase if i dont have a graph.....then the teacher will take marks off.........
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