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mariomintchev

  • 2 years ago

I need help with problem 1 part (d)

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

  3. mariomintchev
    • 2 years ago
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    @AccessDenied @jim_thompson5910 @LoveYou*69 @Luis_Rivera @phi @robtobey

  4. mariomintchev
    • 2 years ago
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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

  5. mariomintchev
    • 2 years ago
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    @satellite73

  6. mariomintchev
    • 2 years ago
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    @Hero @dumbcow

  7. mariomintchev
    • 2 years ago
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    @amistre64 @Jemurray3

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

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

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

  11. amistre64
    • 2 years ago
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    i recall Expected Value is something like np ...

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

  13. amistre64
    • 2 years ago
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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

  14. amistre64
    • 2 years ago
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    thats my understanding of it

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

  16. satellite73
    • 2 years ago
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    oh just part d)

  17. amistre64
    • 2 years ago
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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

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

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

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

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

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

  23. mariomintchev
    • 2 years ago
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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)

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

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

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

  27. mariomintchev
    • 2 years ago
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    1d lol

  28. amistre64
    • 2 years ago
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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

  29. amistre64
    • 2 years ago
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    ir is it divide by number of rows?

  30. amistre64
    • 2 years ago
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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

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

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

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

  34. mariomintchev
    • 2 years ago
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    you didnt answer my question lol

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

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

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

  39. mariomintchev
    • 2 years ago
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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?

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

  41. amistre64
    • 2 years ago
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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? \[Var=\frac{156(11-\mu)^2+110(12-\mu)^2+...+1(19-\mu)^2}{500}\]

  42. amistre64
    • 2 years ago
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    thats better.

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

  44. phi
    • 2 years ago
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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

  45. mariomintchev
    • 2 years ago
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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.

  46. phi
    • 2 years ago
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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

  47. mariomintchev
    • 2 years ago
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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?

  48. phi
    • 2 years ago
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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

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

  50. mariomintchev
    • 2 years ago
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    confused -_-

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

  52. mariomintchev
    • 2 years ago
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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 ?

  53. phi
    • 2 years ago
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    yes,

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

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

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

  57. phi
    • 2 years ago
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    yes

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

  59. mariomintchev
    • 2 years ago
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    not sure if .10 is right though

  60. phi
    • 2 years ago
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    I think you use the binomial distribution

  61. phi
    • 2 years ago
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    http://en.wikipedia.org/wiki/Binomial_distribution

  62. phi
    • 2 years ago
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    does that sound familiar ?

  63. mariomintchev
    • 2 years ago
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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.

  64. phi
    • 2 years ago
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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)

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

  66. phi
    • 2 years ago
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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

  67. phi
    • 2 years ago
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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

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

  69. phi
    • 2 years ago
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    wikipedia gives an example, scroll down...

  70. mariomintchev
    • 2 years ago
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    wait i see it now

  71. mariomintchev
    • 2 years ago
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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???

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

  73. phi
    • 2 years ago
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    question 4 looks like you use the Poisson distribution

  74. phi
    • 2 years ago
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    But I have to go now.

  75. mariomintchev
    • 2 years ago
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    ok thanks for the help!!

  76. phi
    • 2 years ago
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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).

  77. mariomintchev
    • 2 years ago
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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.

  78. mariomintchev
    • 2 years ago
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    @phi

  79. phi
    • 2 years ago
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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

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