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mariomintchev

I need help with problem 1 part (d)

  • one year ago
  • one year ago

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  1. mariomintchev
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    • one year ago
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  2. mariomintchev
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    I'm thinking 2.5 ? ?

    • one year ago
  3. mariomintchev
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    @AccessDenied @jim_thompson5910 @LoveYou*69 @Luis_Rivera @phi @robtobey

    • one year ago
  4. mariomintchev
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    Also, for problem 2 part (a)... what formulas would i use to calculate the problems? E(x+y)=E(x)+E(y) ??? For the expected value and V(x+y)=V(x)+V(y)+2COV (x,y) ??? For the variance

    • one year ago
  5. mariomintchev
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    @satellite73

    • one year ago
  6. mariomintchev
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    @Hero @dumbcow

    • one year ago
  7. mariomintchev
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    @amistre64 @Jemurray3

    • one year ago
  8. mariomintchev
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    i think the answer for my first question that i asked above is 2.96... any help with the second?

    • one year ago
  9. amistre64
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    i cant make any sense of how they presented the information in the graph :/

    • one year ago
  10. mariomintchev
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    can you attempt Q2 Part A? I think I got the other one.

    • one year ago
  11. amistre64
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    i recall Expected Value is something like np ...

    • one year ago
  12. mariomintchev
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    do i just add up em all up and divide by the number of games?

    • one year ago
  13. amistre64
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    it looks to me like the number of games that the number of FreeThrows were made in. so in 105 games, each game had 11 FreeThrows made. giving us 105 * 11 FTs for the first row but what is the expected number of Free Throws with the given data? 105/500 percent of the time they made 11 shots per game. 110/500 percent of the time they made 12 shots per game etc ... then you add up all the EV values that each row represents

    • one year ago
  14. amistre64
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    thats my understanding of it

    • one year ago
  15. satellite73
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    for the first one, don't you add up the column?

    • one year ago
  16. satellite73
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    oh just part d)

    • one year ago
  17. amistre64
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    another thought I have is to just multiply across the rows, that will give you the total number of FTs for those games, then add up all the FTs that were made and divide it by 500 for an average

    • one year ago
  18. mariomintchev
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    for the first one i did: (4.0*.41)+(3.0*.36)+(2.0*.12)+(0.0*.11)=2.96

    • one year ago
  19. mariomintchev
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    but keep going with the other one... it's more confusing...

    • one year ago
  20. amistre64
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    \[EV=avg.=\frac{105(11)+110(12)+167(13)+...+1(19)}{500}\]

    • one year ago
  21. satellite73
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    sorry i am confused which one is it? 1d)? or 2?

    • one year ago
  22. satellite73
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    @amistre64 has it for 2 for sure

    • one year ago
  23. mariomintchev
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    Problem 2 Part A (is what Amistre is helping me with right now) the other one was Problem 1 Part D (which I got 2.96 for)

    • one year ago
  24. amistre64
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    i just noticed when I put my two thoughts together that they are really the same thought ;)

    • one year ago
  25. mariomintchev
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    lol how would i do the variance part of the question?

    • one year ago
  26. satellite73
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    i am not sure your answer to 1b) is right though

    • one year ago
  27. mariomintchev
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    1d lol

    • one year ago
  28. amistre64
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    find the difference that each row value is from the mean, square it, add up all the square and divide by ..... 500 i believe

    • one year ago
  29. amistre64
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    ir is it divide by number of rows?

    • one year ago
  30. amistre64
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    standard deviation is the average deviation that we expect to get huddled about the mean. (105(11) - mean)^2 +(110(12)- mean)^2 + ......................... summed value of squares divide that by 500 to get the variance

    • one year ago
  31. mariomintchev
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    the mean is the # of free throws made divided by 9 right?

    • one year ago
  32. amistre64
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    hmm the mean is prolly a small number 105 * (11 - mean)^2 is prolly more accurate to do

    • one year ago
  33. amistre64
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    might be easier to do in excel fer sure :)

    • one year ago
  34. mariomintchev
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    you didnt answer my question lol

    • one year ago
  35. mariomintchev
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    the mean is the # of free throws made divided by 9 right? <<<<<

    • one year ago
  36. amistre64
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    • one year ago
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  37. amistre64
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    the mean is the number of free throws per game for 500 games divided by 500

    • one year ago
  38. amistre64
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    im still uneasy about the var and sd calculations for some reason.

    • one year ago
  39. mariomintchev
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    wait isnt that what we did for the expected value? 105*11 + 110*12 + ...... / 500 ?? <<That's the mean and expected value?

    • one year ago
  40. amistre64
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    the mean is the expected value; yes. You expect the average results to happen in any given situation.

    • one year ago
  41. amistre64
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    if we had a list of ALL the games in a one to one list: it would look like: 1, 11 2, 11 3, 11 4, 11 ... 156, 11 157, 12 158, 12 .... etc for all 500 games played. if we take this list to find the variance, we have to subtract the mean from each data point, and square the reults (11-mean)^2 (11-mean)^2 (11-mean)^2 ... 156 times (12-mean)^2 (12-mean)^2 (12-mean)^2 ... 110 times and for 500 games. adding up all those squared parts and dividing by 500 gives us variance right? \[Var=\frac{156(11-\mu)^2+110(12-\mu)^2+...+1(19-\mu)^2}{500}\]

    • one year ago
  42. amistre64
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    thats better.

    • one year ago
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  43. mariomintchev
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    For Problem 3 A through C, are all 3 answers .10??

    • one year ago
  44. phi
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    How did ou get 2.96 for question 1 d ? I would multiply the probabilities by these numbers and add them up 4 3.5 3 2 3.5 3 2.5 1.5 3 2.5 2 1 2 1.5 1 0

    • one year ago
  45. mariomintchev
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    (4.0*.41)+(3.0*.36)+(2.0*.12)+(0.0*.11)=2.96 is what i did. i added all the columns up and then multiplied by the different GPAS.

    • one year ago
  46. phi
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    For Q 2 I would get the total number of games and then create a weight # of games/total and use that to weight the number of free throws e.g. for the first row, 105/total * 11 do that for each row, and add up the results

    • one year ago
  47. mariomintchev
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    4 3.5 3 2 3.5 3 2.5 1.5 3 2.5 2 1 2 1.5 1 0 how did you get these numbers?

    • one year ago
  48. phi
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    i added all the columns up and then multiplied by the different GPAS. that does not sound right. The expected value is the sum of the outcome * probability of the outcome. For example, a 4.0 (2 A's) has prob 0.1 multiply to get 0.4 do that for each box, and add up the numbers

    • one year ago
  49. phi
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    how did you get these numbers? those are the gpa for each box

    • one year ago
  50. mariomintchev
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    confused -_-

    • one year ago
  51. phi
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    First, you can figure out what the gpa would be for each little box?

    • one year ago
  52. mariomintchev
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    so you take each intersection, e.g A & B and take their GPAs and average them out? so, 4.0+3.0 / 2 ?

    • one year ago
  53. phi
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    yes,

    • one year ago
  54. phi
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    Each box represents a possible grade assignment, and consequently a particular gpa

    • one year ago
  55. phi
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    the expected value will be each gpa times probability of that gpa all added up

    • one year ago
  56. mariomintchev
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    so for the first one it would be: 4.0 * 0.10?

    • one year ago
  57. phi
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    yes

    • one year ago
  58. mariomintchev
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    can you help with problem 3? i got .10 for A through C

    • one year ago
  59. mariomintchev
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    not sure if .10 is right though

    • one year ago
  60. phi
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    I think you use the binomial distribution

    • one year ago
  61. phi
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    http://en.wikipedia.org/wiki/Binomial_distribution

    • one year ago
  62. phi
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    does that sound familiar ?

    • one year ago
  63. mariomintchev
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    somewhat, yes. i just started the semester recently so i'm not too keen on all of this. i haven't taken a lot of stats courses.

    • one year ago
  64. phi
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    notice part B and part C have to add to 100%. either an odd number or an even number got A's, and the two conditions have to add to probability 1: It is 100% certain either an odd number or an even # got an A (counting zero as even)

    • one year ago
  65. phi
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    for part A, I think you use \[\Pr(k \text{ occurrences }) = \left(\begin{matrix}n \\ k\end{matrix}\right)p^k (1-p)^{n-k}\]

    • one year ago
  66. phi
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    p is the probability of an A 0.05 1-p= 0.95 n is 20 we have to do this computation 3 times, for k=0,1, 2 ) and add up the results k is the # of A's in the group of 20

    • one year ago
  67. phi
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    I think for part C you do it brute force for k= 1,3,5,7,9 part B will be 1 - part C

    • one year ago
  68. mariomintchev
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    can you fill in the formula for part A so I can see it

    • one year ago
  69. phi
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    wikipedia gives an example, scroll down...

    • one year ago
  70. mariomintchev
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    wait i see it now

    • one year ago
  71. mariomintchev
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    so it would be .05^0 * (1-.05) ^20-0 for 0 and then replace the zero with a 1 and then 2???

    • one year ago
  72. phi
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    there is an n choose k out front or 20 choose 0 (which I think is 1) but yes

    • one year ago
  73. phi
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    question 4 looks like you use the Poisson distribution

    • one year ago
  74. phi
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    But I have to go now.

    • one year ago
  75. mariomintchev
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    ok thanks for the help!!

    • one year ago
  76. phi
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    For problem 3 I hope you caught my mistake. p (chance of an A) is 20% so p= 20/100= 0.2 and (1-p) is 0.8 (somehow I got 0.05 which is wrong).

    • one year ago
  77. mariomintchev
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    o my god. now i gotta go back and re do everything. ahhhh. haha can you help with problem 4 while re calculate everything from prob 3. i told you what i was having issues with in the message i sent you yesterday after you left.

    • one year ago
  78. mariomintchev
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    @phi

    • one year ago
  79. phi
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    for problem 4, use \[ Pr(k)= f(k,\lambda) =\frac{λ^k e^{-λ}}{k!} \] where λ is the expected number of events in an interval and k is the number of events that occur for part (a) λ is 2.5 for part (b) λ is 2.5*8= 20

    • one year ago
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