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mariomintchev Group Title

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

  • one year ago
  • one year ago

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

    • one year ago
  3. mariomintchev Group Title
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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???

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

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

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

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

    • one year ago
  8. dumbcow Group Title
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    haha :)

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

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

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

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

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

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

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

    • one year ago
  16. mariomintchev Group Title
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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 ???

    • one year ago
  17. dumbcow Group Title
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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

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

    • one year ago
  19. dumbcow Group Title
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    right

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

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

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

    • one year ago
  23. mariomintchev Group Title
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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???

    • one year ago
  24. mariomintchev Group Title
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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

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

    • one year ago
  26. dumbcow Group Title
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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

    • one year ago
  27. mariomintchev Group Title
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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 )

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

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

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

    • one year ago
  31. mariomintchev Group Title
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    number 7

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

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

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

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

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

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

    • one year ago
  38. satellite73 Group Title
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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

    • one year ago
  39. mariomintchev Group Title
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    ok yeah

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

    • one year ago
  41. satellite73 Group Title
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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 \(.25\)

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

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

    • one year ago
  44. satellite73 Group Title
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    no

    • one year ago
  45. mariomintchev Group Title
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    *arent

    • one year ago
  46. satellite73 Group Title
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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

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

    • one year ago
  48. satellite73 Group Title
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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 \[0.59870633-0.2266273\]

    • one year ago
  49. satellite73 Group Title
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    you got that?

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

    • one year ago
  51. satellite73 Group Title
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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

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

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

    • one year ago
  54. satellite73 Group Title
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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

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

    • one year ago
  56. satellite73 Group Title
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    ok good

    • one year ago
  57. satellite73 Group Title
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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

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

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

    • one year ago
  60. satellite73 Group Title
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    yeah that looks right

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

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

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

    • one year ago
  64. satellite73 Group Title
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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%

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

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

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

    • one year ago
  68. satellite73 Group Title
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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

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

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

    • one year ago
  71. mariomintchev Group Title
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    .82894

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

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

    • one year ago
  74. satellite73 Group Title
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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

    • one year ago
  75. satellite73 Group Title
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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 \[x=1.64\times 20+180\]

    • one year ago
  76. mariomintchev Group Title
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    ...ok

    • one year ago
  77. mariomintchev Group Title
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    and then what?

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

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

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

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

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

    • one year ago
  84. satellite73 Group Title
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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

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

    • one year ago
  86. satellite73 Group Title
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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

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

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

    • one year ago
  89. mariomintchev Group Title
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    thanks a lot!

    • one year ago
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