## mariomintchev Group Title I need help with problem 1 part (d) one year ago one year ago

1. mariomintchev Group Title

2. mariomintchev Group Title

I'm thinking 2.5 ? ?

3. mariomintchev Group Title

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

4. mariomintchev Group Title

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 Group Title

@satellite73

6. mariomintchev Group Title

@Hero @dumbcow

7. mariomintchev Group Title

@amistre64 @Jemurray3

8. mariomintchev Group Title

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

9. amistre64 Group Title

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

10. mariomintchev Group Title

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

11. amistre64 Group Title

i recall Expected Value is something like np ...

12. mariomintchev Group Title

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

13. amistre64 Group Title

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 Group Title

thats my understanding of it

15. satellite73 Group Title

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

16. satellite73 Group Title

oh just part d)

17. amistre64 Group Title

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 Group Title

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

19. mariomintchev Group Title

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

20. amistre64 Group Title

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

21. satellite73 Group Title

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

22. satellite73 Group Title

@amistre64 has it for 2 for sure

23. mariomintchev Group Title

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 Group Title

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

25. mariomintchev Group Title

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

26. satellite73 Group Title

27. mariomintchev Group Title

1d lol

28. amistre64 Group Title

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 Group Title

ir is it divide by number of rows?

30. amistre64 Group Title

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 Group Title

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

32. amistre64 Group Title

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

33. amistre64 Group Title

might be easier to do in excel fer sure :)

34. mariomintchev Group Title

you didnt answer my question lol

35. mariomintchev Group Title

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

36. amistre64 Group Title

37. amistre64 Group Title

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

38. amistre64 Group Title

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

39. mariomintchev Group Title

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

40. amistre64 Group Title

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

41. amistre64 Group Title

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 Group Title

thats better.

43. mariomintchev Group Title

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

44. phi Group Title

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 Group Title

(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 Group Title

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 Group Title

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 Group Title

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 Group Title

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

50. mariomintchev Group Title

confused -_-

51. phi Group Title

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

52. mariomintchev Group Title

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 Group Title

yes,

54. phi Group Title

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

55. phi Group Title

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

56. mariomintchev Group Title

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

57. phi Group Title

yes

58. mariomintchev Group Title

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

59. mariomintchev Group Title

not sure if .10 is right though

60. phi Group Title

I think you use the binomial distribution

61. phi Group Title
62. phi Group Title

does that sound familiar ?

63. mariomintchev Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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

69. phi Group Title

wikipedia gives an example, scroll down...

70. mariomintchev Group Title

wait i see it now

71. mariomintchev Group Title

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 Group Title

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

73. phi Group Title

question 4 looks like you use the Poisson distribution

74. phi Group Title

But I have to go now.

75. mariomintchev Group Title

ok thanks for the help!!

76. phi Group Title

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 Group Title

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 Group Title

@phi

79. phi Group Title

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