## mariomintchev 2 years ago I need to know if I use the Poisson Distribution for problem 5 (A-C).

1. mariomintchev

2. dumbcow

No use the Binomial distribution p= 1/3 for tie p=1/6 for "0-0" tie

3. mariomintchev

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???

4. mariomintchev

@jim_thompson5910 @karatechopper @JuanitaM

5. dumbcow
6. dumbcow

n=10 and k=3

7. mariomintchev

so im correct. woohoo! :D

8. dumbcow

haha :)

9. mariomintchev

ok, so i get .0021676912. is that the final answer or is there anything else i needa do?

10. dumbcow

for part A ? i get about 0.26

11. mariomintchev

yeah, how'd u get .26 ??

12. dumbcow

just plugged numbers in formula ... well i cheated and used Excel :) p = 1/3 n=10 k=3 $\rightarrow \left(\begin{matrix}10 \\ 3\end{matrix}\right)*(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = 120*\frac{128}{3^{10}} = 0.2601$

13. mariomintchev

im confused about the second half... after the first equal sign

14. dumbcow

$\left(\begin{matrix}10 \\ 3\end{matrix}\right) = \frac{10!}{7! 3!} = 120$ $(\frac{1}{3})^{3} *(\frac{2}{3})^{7} = \frac{1}{3^{3}}*\frac{2^{7}}{3^{7}} = \frac{2^{7}}{3^{10}} = \frac{128}{59,049}$

15. mariomintchev

ok. so i was getting .0021676912 because i didn't multiply it by 120. makes sense.

16. mariomintchev

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 ???

17. dumbcow

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

18. mariomintchev

o because half would equal zero-zero ??

19. dumbcow

right

20. mariomintchev

and part Cs answer will basically be the same as part B since they're both p=1/6?

21. dumbcow

umm yep thats what it looks like

22. mariomintchev

ok coo ! :)

23. mariomintchev

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???

24. mariomintchev

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

25. mariomintchev

|dw:1360115375877:dw|

26. dumbcow

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

27. mariomintchev

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 )

28. mariomintchev

@satellite73 can you continue where @dumbcow left off??? they are no longer online. :(

29. satellite73

4 is poisson, 5 is binomial

30. satellite73

what number are you on?

31. mariomintchev

number 7

32. mariomintchev

it's asking me to repeat the steps of #6 but idk what to really do

33. satellite73

damn i don't know squat about the normal distribution sorry

34. mariomintchev

its like the picture i drew above ^^^

35. satellite73

i know, but you need a table to calculate these

36. satellite73

the mean is 180 right?

37. mariomintchev

yes and the standard deviation is 20

38. satellite73

then you need to convert to a z score or something right? so you can look these up in a table

39. mariomintchev

ok yeah

40. mariomintchev
41. satellite73

for example $\frac{165-180}{20}=-.75$

42. satellite73

and $\frac{185-180}{20}=.25$ so you need to use a table to find the area under the normal curve between $$-.75$$ and $$.25$$

43. satellite73

you can use this http://www.danielsoper.com/statcalc3/calc.aspx?id=2

44. mariomintchev

i dont think thats how it works... are both sides supposed to be equal but opposite?

45. satellite73

no

46. mariomintchev

*arent

47. satellite73

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

48. satellite73

so your job is to find the area under the curve from -.75 to .25 this takes a couple steps

49. satellite73

from the table , to the left of .25 is 0.59870633 and to the left of -.75 is 0.2266273 so you use $0.59870633-0.2266273$

50. satellite73

you got that?

51. mariomintchev

im soaking it all in

52. satellite73

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

53. satellite73

|dw:1360121214382:dw|

54. satellite73

|dw:1360121361823:dw|

55. satellite73

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

56. mariomintchev

yeah i see what you are doing

57. satellite73

ok good

58. satellite73

site is freezing up on me, but you can use the calculator i linked to find the z scores, then the area you want

59. mariomintchev

ok so we get 0.37207903 or 37% chance ?

60. satellite73

i am having trouble scrolling up, but i did find the numbers you need, i just didn't subtract

61. satellite73

yeah that looks right

62. mariomintchev

can you assist me with #8 too?

63. mariomintchev

|dw:1360121910130:dw|

64. mariomintchev

where do i go from here??

65. satellite73

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%

66. satellite73

do you have a table?

67. satellite73

oh forget what i wrote above you just need to go to 95% short people can sit anyway

68. mariomintchev

im confused. can you reword.

69. satellite73

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

70. mariomintchev
71. satellite73

yeah we can use that

72. mariomintchev

.82894

73. satellite73

so we find .95 in the table, and see what the corresponding z score is

74. satellite73

looks like about 1.64 or 1.65 half way between them actually lets pick one

75. satellite73

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

76. satellite73

remember we computed the z score for say $$x$$ by $\frac{x-180}{20}$ so if $\frac{x-180}{20}=1.64$ then $x=1.64\times 20+180$

77. mariomintchev

...ok

78. mariomintchev

and then what?

79. mariomintchev

80. satellite73

yeah i forgot what the question was can you repost so i don't have to keep scrolling down?

81. mariomintchev

82. satellite73

nvm the limit of the height adjustment is what you wanted so yes, that should be the answer

83. mariomintchev

should i solve or leave as is

84. satellite73

of course they want a number i would say about 213 or so

85. satellite73

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

86. mariomintchev

i got 221 as my final answer

87. satellite73

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

88. mariomintchev

for the 98 % curve

89. satellite73

yeah i go the same thing gotta run good luck!

90. mariomintchev

thanks a lot!