## mariomintchev 2 years ago I need help with problem 1 part (d)

1. mariomintchev

2. mariomintchev

I'm thinking 2.5 ? ?

3. mariomintchev

@AccessDenied @jim_thompson5910 @LoveYou*69 @Luis_Rivera @phi @robtobey

4. mariomintchev

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

@satellite73

6. mariomintchev

@Hero @dumbcow

7. mariomintchev

@amistre64 @Jemurray3

8. mariomintchev

i think the answer for my first question that i asked above is 2.96... any help with the second?

9. amistre64

i cant make any sense of how they presented the information in the graph :/

10. mariomintchev

can you attempt Q2 Part A? I think I got the other one.

11. amistre64

i recall Expected Value is something like np ...

12. mariomintchev

do i just add up em all up and divide by the number of games?

13. amistre64

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

thats my understanding of it

15. satellite73

for the first one, don't you add up the column?

16. satellite73

oh just part d)

17. amistre64

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

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

19. mariomintchev

but keep going with the other one... it's more confusing...

20. amistre64

$EV=avg.=\frac{105(11)+110(12)+167(13)+...+1(19)}{500}$

21. satellite73

sorry i am confused which one is it? 1d)? or 2?

22. satellite73

@amistre64 has it for 2 for sure

23. mariomintchev

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

i just noticed when I put my two thoughts together that they are really the same thought ;)

25. mariomintchev

lol how would i do the variance part of the question?

26. satellite73

27. mariomintchev

1d lol

28. amistre64

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

ir is it divide by number of rows?

30. amistre64

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

the mean is the # of free throws made divided by 9 right?

32. amistre64

hmm the mean is prolly a small number 105 * (11 - mean)^2 is prolly more accurate to do

33. amistre64

might be easier to do in excel fer sure :)

34. mariomintchev

you didnt answer my question lol

35. mariomintchev

the mean is the # of free throws made divided by 9 right? <<<<<

36. amistre64

37. amistre64

the mean is the number of free throws per game for 500 games divided by 500

38. amistre64

im still uneasy about the var and sd calculations for some reason.

39. mariomintchev

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

40. amistre64

the mean is the expected value; yes. You expect the average results to happen in any given situation.

41. amistre64

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

thats better.

43. mariomintchev

For Problem 3 A through C, are all 3 answers .10??

44. phi

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

(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

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

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

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

how did you get these numbers? those are the gpa for each box

50. mariomintchev

confused -_-

51. phi

First, you can figure out what the gpa would be for each little box?

52. mariomintchev

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

yes,

54. phi

Each box represents a possible grade assignment, and consequently a particular gpa

55. phi

the expected value will be each gpa times probability of that gpa all added up

56. mariomintchev

so for the first one it would be: 4.0 * 0.10?

57. phi

yes

58. mariomintchev

can you help with problem 3? i got .10 for A through C

59. mariomintchev

not sure if .10 is right though

60. phi

I think you use the binomial distribution

61. phi
62. phi

does that sound familiar ?

63. mariomintchev

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

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

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

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

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

can you fill in the formula for part A so I can see it

69. phi

wikipedia gives an example, scroll down...

70. mariomintchev

wait i see it now

71. mariomintchev

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

there is an n choose k out front or 20 choose 0 (which I think is 1) but yes

73. phi

question 4 looks like you use the Poisson distribution

74. phi

But I have to go now.

75. mariomintchev

ok thanks for the help!!

76. phi

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

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

@phi

79. phi

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