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mariomintchev

  • one year ago

I need to know if I use the Poisson Distribution for problem 5 (A-C).

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  1. mariomintchev
    • one year ago
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  2. dumbcow
    • one year ago
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    No use the Binomial distribution p= 1/3 for tie p=1/6 for "0-0" tie

  3. mariomintchev
    • one year ago
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    so i would use this formula: P^k * (1-p)^n-k and for part A, I would plug in 10 for n and 3 for k???

  4. mariomintchev
    • one year ago
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    @jim_thompson5910 @karatechopper @JuanitaM

  5. dumbcow
    • one year ago
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    http://en.wikipedia.org/wiki/Binomial_distribution#Probability_mass_function

  6. dumbcow
    • one year ago
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    n=10 and k=3

  7. mariomintchev
    • one year ago
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    so im correct. woohoo! :D

  8. dumbcow
    • one year ago
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    haha :)

  9. mariomintchev
    • one year ago
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    ok, so i get .0021676912. is that the final answer or is there anything else i needa do?

  10. dumbcow
    • one year ago
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    for part A ? i get about 0.26

  11. mariomintchev
    • one year ago
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    yeah, how'd u get .26 ??

  12. dumbcow
    • one year ago
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    just plugged numbers in formula ... well i cheated and used Excel :) p = 1/3 n=10 k=3 \[\rightarrow \left(\begin{matrix}10 \\ 3\end{matrix}\right)*(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = 120*\frac{128}{3^{10}} = 0.2601\]

  13. mariomintchev
    • one year ago
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    im confused about the second half... after the first equal sign

  14. dumbcow
    • one year ago
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    \[\left(\begin{matrix}10 \\ 3\end{matrix}\right) = \frac{10!}{7! 3!} = 120\] \[(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = \frac{1}{3^{3}}*\frac{2^{7}}{3^{7}} = \frac{2^{7}}{3^{10}} = \frac{128}{59,049}\]

  15. mariomintchev
    • one year ago
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    ok. so i was getting .0021676912 because i didn't multiply it by 120. makes sense.

  16. mariomintchev
    • one year ago
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    so for part B i would just change p to 1/6 ??? (why 1/6 again?) and for part C i would do 1-part B ???

  17. dumbcow
    • one year ago
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    its 1/6 because its half of 1/3 --> 1/3 * 1/2 = 1/6 part C is looking at chance you tie but Not 0-0 --> 1/3 - 1/6 = 1/6

  18. mariomintchev
    • one year ago
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    o because half would equal zero-zero ??

  19. dumbcow
    • one year ago
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    right

  20. mariomintchev
    • one year ago
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    and part Cs answer will basically be the same as part B since they're both p=1/6?

  21. dumbcow
    • one year ago
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    umm yep thats what it looks like

  22. mariomintchev
    • one year ago
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    ok coo ! :)

  23. mariomintchev
    • one year ago
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    alright, so im on #7 and it's wanting the same info as #6 but normally distributed... would i use : f(x)= ( 1 / [sigma * sqrt(2pi)] ) * e ^ (x-mu/sigma) to plug in the same info from 6???

  24. mariomintchev
    • one year ago
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    here are the answers to #6: a) picture b) 1/(b-a) = 1/(210-150) = 1/60 c) (160-150)/(210-150) = 10/60 = 1/6 d) 1/6 * 1/6 = 1/36 e) (185-165)/(210-150) = 20/60 = 2/6 f) (210-200)/(210-150) = 10/60 = 1/6

  25. mariomintchev
    • one year ago
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    |dw:1360115375877:dw|

  26. dumbcow
    • one year ago
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    yes that is the function for normal distribution but since its complex....you need to use probability table for normal distribution problems you have to determine z-score (standardized value) first anyway, check your answers from #6 again first though the mean has to be a number between 150 and 210 .... not 1/60

  27. mariomintchev
    • one year ago
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    i got 1/60 by using the formula but i guess thats not correct since it makes no sense to have a mean of 1/60. i guess the mean is 180 ( 150+210 / 2 )

  28. mariomintchev
    • one year ago
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    @satellite73 can you continue where @dumbcow left off??? they are no longer online. :(

  29. satellite73
    • one year ago
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    4 is poisson, 5 is binomial

  30. satellite73
    • one year ago
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    what number are you on?

  31. mariomintchev
    • one year ago
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    number 7

  32. mariomintchev
    • one year ago
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    it's asking me to repeat the steps of #6 but idk what to really do

  33. satellite73
    • one year ago
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    damn i don't know squat about the normal distribution sorry

  34. mariomintchev
    • one year ago
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    its like the picture i drew above ^^^

  35. satellite73
    • one year ago
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    i know, but you need a table to calculate these

  36. satellite73
    • one year ago
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    the mean is 180 right?

  37. mariomintchev
    • one year ago
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    yes and the standard deviation is 20

  38. satellite73
    • one year ago
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    then you need to convert to a z score or something right? so you can look these up in a table

  39. mariomintchev
    • one year ago
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    ok yeah

  40. satellite73
    • one year ago
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    for example \[\frac{165-180}{20}=-.75\]

  41. satellite73
    • one year ago
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    and \[\frac{185-180}{20}=.25\] so you need to use a table to find the area under the normal curve between \(-.75\) and \(.25\)

  42. satellite73
    • one year ago
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    you can use this http://www.danielsoper.com/statcalc3/calc.aspx?id=2

  43. mariomintchev
    • one year ago
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    i dont think thats how it works... are both sides supposed to be equal but opposite?

  44. satellite73
    • one year ago
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    no

  45. mariomintchev
    • one year ago
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    *arent

  46. satellite73
    • one year ago
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    not unless they happen to be your question is between 165 and 185 which is not symmetric about 180 it is 5 up and 10 down

  47. satellite73
    • one year ago
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    so your job is to find the area under the curve from -.75 to .25 this takes a couple steps

  48. satellite73
    • one year ago
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    from the table , to the left of .25 is 0.59870633 and to the left of -.75 is 0.2266273 so you use \[0.59870633-0.2266273\]

  49. satellite73
    • one year ago
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    you got that?

  50. mariomintchev
    • one year ago
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    im soaking it all in

  51. satellite73
    • one year ago
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    the normal curve is symmetric the way the tables work is that they give you the area under the curve to the left of your z score so if you want the area between two z scores you take the larger z score, find the area in a table, then take the smaller area, and subtract

  52. satellite73
    • one year ago
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    |dw:1360121214382:dw|

  53. satellite73
    • one year ago
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    |dw:1360121361823:dw|

  54. satellite73
    • one year ago
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    area between -.75 and .25 is what you want it is the area to the left of .25 minus the area to the left of -.75

  55. mariomintchev
    • one year ago
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    yeah i see what you are doing

  56. satellite73
    • one year ago
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    ok good

  57. satellite73
    • one year ago
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    site is freezing up on me, but you can use the calculator i linked to find the z scores, then the area you want

  58. mariomintchev
    • one year ago
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    ok so we get 0.37207903 or 37% chance ?

  59. satellite73
    • one year ago
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    i am having trouble scrolling up, but i did find the numbers you need, i just didn't subtract

  60. satellite73
    • one year ago
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    yeah that looks right

  61. mariomintchev
    • one year ago
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    can you assist me with #8 too?

  62. mariomintchev
    • one year ago
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    |dw:1360121910130:dw|

  63. mariomintchev
    • one year ago
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    where do i go from here??

  64. satellite73
    • one year ago
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    we got to look in a table and see what the right hand point give 97.5 % of the area (so the left hand point will have 2.5% and the area between them will be 95%

  65. satellite73
    • one year ago
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    do you have a table?

  66. satellite73
    • one year ago
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    oh forget what i wrote above you just need to go to 95% short people can sit anyway

  67. mariomintchev
    • one year ago
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    im confused. can you reword.

  68. satellite73
    • one year ago
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    i think 95% of the people lie within two standard deviations of the mean, meaning if you look at the table you should see for .95 a z score of 2 you have to check this because i do not have a table if this is true, then two standard deviations is 40 and 180 + 40 = 120 is the max height

  69. mariomintchev
    • one year ago
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    are talking about this table? http://assets.openstudy.com/updates/attachments/5061f687e4b0583d5cd2e54c-mahlatse1-1348604740357-standardnormaltable.pdf

  70. satellite73
    • one year ago
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    yeah we can use that

  71. mariomintchev
    • one year ago
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    .82894

  72. satellite73
    • one year ago
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    so we find .95 in the table, and see what the corresponding z score is

  73. satellite73
    • one year ago
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    looks like about 1.64 or 1.65 half way between them actually lets pick one

  74. satellite73
    • one year ago
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    for the sake of argument, lets say the z score is 1.64 then we convert back to height \[1.64\times 20+180\] will give what we want

  75. satellite73
    • one year ago
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    remember we computed the z score for say \(x\) by \[\frac{x-180}{20}\] so if \[\frac{x-180}{20}=1.64\] then \[x=1.64\times 20+180\]

  76. mariomintchev
    • one year ago
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    ...ok

  77. mariomintchev
    • one year ago
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    and then what?

  78. mariomintchev
    • one year ago
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    thats the final answer??

  79. satellite73
    • one year ago
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    yeah i forgot what the question was can you repost so i don't have to keep scrolling down?

  80. mariomintchev
    • one year ago
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  81. satellite73
    • one year ago
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    nvm the limit of the height adjustment is what you wanted so yes, that should be the answer

  82. mariomintchev
    • one year ago
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    should i solve or leave as is

  83. satellite73
    • one year ago
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    of course they want a number i would say about 213 or so

  84. satellite73
    • one year ago
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    the next one is similar find .98 in the body of the table, find the corresponding z score and work backwards to find the height

  85. mariomintchev
    • one year ago
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    i got 221 as my final answer

  86. satellite73
    • one year ago
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    looks like a z score of about 2.05 there is nothing really profound about this, it just says that 98% are less than 2.05 standard deviations above the mean since the standard deviation is 20 and the mean is 180 that means 98% are under \(2.05\times 20+180=221\) cm tall

  87. mariomintchev
    • one year ago
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    for the 98 % curve

  88. satellite73
    • one year ago
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    yeah i go the same thing gotta run good luck!

  89. mariomintchev
    • one year ago
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    thanks a lot!

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